Let $T = \omega \cap \Gamma, D = XY \cap BC, E = AM \cap TN',$ and $A_1 = BB \cap CC$. Let $N'$ be the point on $\Gamma$ diametrically opposite to $N.$ Let $\gamma$ be the circle centered at $K$ with radius $KP.$ We'll first show that $\gamma$ is tangent to $\Gamma.$ Observe that since $\angle ZTN = \angle N'TN = 90$, we have that $Z \in TN'.$ Also, we have that $DZ^2 = DX \cdot DY = DB \cdot DC.$ Lemma 1. $DT$ is tangent to $\Gamma$. Proof. Since $\angle BTZ = \angle CTZ$, we conclude that $Z, T$ are both on the $T-$Apollonius circle of $\triangle BTC.$ Suppose that $D' \in BC$ is the center of the $T-$Apollonius circle of $\triangle BTC.$ Since it's well-known that this circle is orthogonal to $(\triangle BTC) = \Gamma$, we have that $D'Z = D'B \cdot D'C \Rightarrow D' = D.$ So we have that $DT = DZ.$ Hence, $\angle N'TD = \angle ZTD = \angle DZT = \angle BZT = \angle N'BC + \angle BN'T = \angle N'CB + \angle BCT = \angle N'CT$. This implies that $DT$ is tangent to $\Gamma.$ $\blacksquare$ Lemma 2. $MK || NT.$ Proof. Observe that from Lemma $1$, we know that $pol_{\omega} (D) = TA_1$, where $pol_{c} (\alpha)$ denotes the polar (resp. pole) of point (resp. line) $\alpha$ w.r.t. circle $c$. Therefore, we have that $TA_1 \perp OD.$ We also have by symmetry w.r.t. $OD$ that $AM \perp OD$, and so hence $AM || TA_1.$ Thus, as $TA_1$ is the $T-$symmedian of $\triangle BTC$, $TM$ is the $T-$median of $\triangle BTC$, and $TN$ is the $T-$external angle-bisector of $\triangle BTC$, we get that the directions of $TM, AM$ are symmetric w.r.t. $TN$ (i.e., we can pick a line parallel to $TN$ so that if $AM$ is reflect over it, it becomes $TM$). Since $KN || AM$, we have that $\angle KNT = \angle NTA_1 = \angle NTM$, which implies that $KMNT$ is an isosceles trapezoid with $MK || NT.$ The lemma is proven. $\blacksquare$ From Lemma $2$, we can finish easily. Indeed, by symmetry w.r.t the perpendicular bisector of $NT$, we know that $\gamma$ is tangent to $\Gamma$ if and only if the circle centered at $M$ with radius $MN$ is tangent to $\Gamma$. However this is obviously true, and so $\gamma$ is also tangent to $\Gamma.$ We will now proceed with the other two parts of the problem (which are clearly symmetric). Lemma 3. $E \in \gamma.$ Proof. We obtained that there exists a line parallel to $TN$ so that $AM, TM$ are reflections over it. Since $TE \perp TN$, this yields that $\triangle TME$ is isosceles with $\angle TEM = \angle MTE.$ Therefore, we have that $\angle TEP = \angle MTE = \angle MNZ = 90 - \angle MZN = 90 - \angle KTP.$ This implies that $E \in \gamma$. $\blacksquare$ Now, orient the triangle $\triangle ABC$ so that $BC$ is "horizontal" and $A$ is "above" $BC.$ Lemma 4. In this way, $E$ is the "highest" point of $\gamma.$ Proof. It suffices to prove that $EK \perp BC.$ We have that $\angle MEK = \angle KPE = 90 - \angle PTE = 90 - \angle EMZ$. This yields that $EK \perp BC$. $\blacksquare$ Observe that there is only one circle which is internally tangent to $\Gamma$ at $T$ and which has "highest" point on $AM$. This circle is $\gamma$, by Lemma $4$ and the first part of the problem (which we have solved). Let's now consider the circle $\Omega$ which is tangent to $\Gamma, \omega_b,$ and $\omega_c$, which touches the arc $BC$ not containing $A.$ Let $\Omega$ touch $\omega_b, \omega_c$ at $F, G$ respectively. Let $E'$ be the "highest" point of $\Omega.$ It suffices to show that $E' \in AM$ and $\Omega$ touches $\Gamma$ at $T.$ Lemma 5. $E' \in AM.$ Proof. Observe that $E', F, B$ are collinear because $F$ is the insimilicenter of $\Omega, \omega_b.$ Analogously $E', G, C$ are collinear. By Monge's Theorem on $\Omega, \omega_b, \omega_c$, we have that $D, F, G$ are collinear. It's now easy to see that $DF \cdot DG = DB \cdot DC$, and we therefore have that $BFGC$ is cyclic. Therefore, by Power of the Point at $E'$, we have that $E'F \cdot E'B = E'G \cdot E'C.$ This implies that $E'$ is on the radical axis of $\omega_b, \omega_c$, which is just $AM.$ $\blacksquare$ Lemma 6. $\Omega$ touches $\Gamma$ at $T.$ Proof. Suppose that $\Omega$ touches $\Gamma$ at $T'$. As seen in the proof of Lemma $5$, we have that $DF \cdot DG = DB \cdot DG$. This implies that $D$ is on the radical axis of $\Omega, \Gamma.$ Since the radical axis of $\Omega, \Gamma$ is just the tangent to $\Gamma$ at $T'$, we have that $DT'$ is tangent to $\Gamma.$ Since we picked $\Omega$ so that $T'$ is on the arc $BC$ not containing $A$, Lemma $1$ implies that $T = T'$ and the lemma is proven. $\blacksquare$ By Lemmas $5, 6$, we have determined that $\Omega$ must be the unique circle which is internally tangent to $\Gamma$ at $T$ and which has "highest" point on $AM$. Hence, we have that $\Omega, \gamma$ are actually the same circle, and so $\gamma$ is tangent to $\omega_b, \omega_c$ as desired. $\square$

Stage 1 Claim 1-We need to prove that the circle with centre $K,$ radius $KP $ is tangent to $\Gamma.$ Proof- Let $\odot(O) $ be the circumcentre of $\triangle ABC,\odot (ZMN)$ cuts $\odot(O)$ at $N $ and $R $ respectively. Note, $XY\cap BC=I, OI\perp XB\implies OI\perp AM.$ Also, $MO\perp BC\implies\measuredangle OIM=\measuredangle OMA=\measuredangle NMP=\measuredangle MNR=\measuredangle MRK.$ Let $R'$ be a point in minor arc $\widehat {BC}$ of $\odot(O)$ such that $R'$ touches $\odot (O)\implies OMR'I$ is a concyclic quadrilateral. $\measuredangle MIO=\measuredangle MR'O=\measuredangle MRK. $ Then $IR'^2=IB\cdot IC=XI\cdot YI=ZI^2\implies \triangle IR'Z $ is isosceles at $(I).$ $\measuredangle R'ON=\measuredangle R'IZ=\measuredangle R'NO=\measuredangle R'ZI\implies R'\in\odot(MNZ)\implies R'=R.$ Hence, $\measuredangle MRK=\measuredangle MRO,$ then conclude $\overline {O,R,K}.$ Then $RK=MN=KP.$ Stage 2 Claim 2- We will show that circle with centre $K, $ radius $KP $ is tangent to $\omega_b, \omega_c$ analogously. Proof- Let $\odot(O_1) $ be the centre of $\omega_b, J\in AM $ such that $KJ\parallel MN, \mathcal {C'}$ is the reflection of $\mathcal {C}$ wrt $O.$ Thus, $\overline {\mathcal {C}, O,B}. $ Because $JP\parallel BX, KJ\parallel MN\parallel O_1B, $ and $KJMN $ is a parallelogram $\implies KJ=MN=KP. $ Note, $\triangle KJP $ and $\triangle O_1BX $ have parallel sides $\implies O_1K, XP, JB $ are concurrent at $S. $ $\measuredangle ZRP=\measuredangle ZKP=180^{\circ}, \measuredangle ZKN=90^{\circ}\implies AM\perp KZ\implies \measuredangle ZRP=180^{\circ}-\left(90^{\circ}+\measuredangle KPJ\right)=90^{\circ}-\measuredangle KPJ.$ $K $ is the circumcentre of $\triangle RPJ\implies \measuredangle JRP=90^{\circ}-\measuredangle KPJ, \overline {Z, J, R}. $ Let $\odot (BMP) $ intersects $\odot(K,KP) $ at $P $ and $S'$ respecrively. Hence, $\measuredangle S'PM=\measuredangle S'PJ=\measuredangle S'RJ=\measuredangle S'RZ=\measuredangle S'BM.$ Then $S', B, R, Z $ are concyclic. Consider $3$ circles $\odot(K, KP),\odot (BMP),\odot(S'BRZ)$ have 3 radical axes $S'B, ZR, MP\implies\overline {S',B,J}. $ Let $S'P $ cuts $\odot(O_1) $ at $S'$ and $X'$ respectively. $\measuredangle S'X'B=\measuredangle S'BM=\measuredangle S'PM\implies BX'\parallel JP\implies X'=X, $ so $S'=S. $ $(K, KP) $ touches $\odot (O_1)$ at $S. $

Complicated this way too much Denote the circle with centre $K$ and radius $KP$ as $\Omega$. Let $O_B$ and $O_C$ denote the circumcircles of $\omega_B$ and $\omega_C$ respectively. Let $S\equiv O_BO_C \cap BC$. Note that $O_B,O_C$ passes through the circumcenter $O$ of $\triangle ABC$. Note that $\{X,B\}$ and $\{Y,C\}$ are symmetric about $O_BO_C$. Next let $T \in \Gamma$ such that $ST$ is tangent to $\Gamma$ and $\{A,T\}$ lie on opposite sides of $BC$. Note that we have by power of point : $$SZ^2= SX \cdot SY = ST^2$$This means that $Z$ is the point on $BC$ where the $T$-appolonius circle of $\triangle BTC$ hits $BC$. Since $Z$ lies in the interior of $BC$, hence $TZ$ bisects $\angle BTC$. Hence $TZ$ passes through $N'$, the antipode of $N$ in $\Gamma$. Next note that $\angle NMZ=90^{\circ}$ implies that $ZN$ is a diameter of $\omega$. Next we have : $$\angle NTZ = \angle NTN' = 90^{\circ} \implies T\in \omega$$ Let $AM$ hit $TZ$ at $R$. Let $\odot (PRT) = \Omega'$ We will now prove that $ST$ is tangent to $\Omega$. This is just angle chasing : $$\angle RPT = \angle MPT = \angle SZT = \angle STR \implies ST \text { is tangent to } \Omega' $$ We now claim that $K$ is the circumcenter of $\Omega'$, hence proving $\Omega=\Omega'$

Note that we have $$\angle MRT = \angle OTM + \angle OTN' = \angle OSM + \angle ON'T = \angle (\perp OS , \perp SM)+ \angle ON'T = \angle OMA + \angle ON'T = \angle MTR \implies \boxed {MT=MR}$$ Let $NT$ hit $AM$ at $R_1$. Then note that $$\angle RTR_1=90^{\circ} \quad \text{and} MR=MT \implies MR=MT=MR_1$$ So we have : $$\angle MKN=\angle MTN = \angle MTR_1=\angle MR_1T=\angle KNT \implies MK \parallel NT \implies \boxed{MK \perp RT}$$ Next note that $\angle ZPR= \angle ZTM=\angle ZRP \implies \boxed{ZR=ZP}$. Also we have $ZK\perp KN \implies ZK \perp RK$. Hence $K$ is the centre of $\Omega'$ and the claim is proven. This proves that $\Omega$ and $\Gamma$ are tangent. Next we will prove that $R$ lies on the radical axis of the circles $\{\omega_B,\odot(S,SZ)\}$ and $\{\omega_C,\odot(S,SZ)\}$. Note that if $\odot(S,SZ)\cap BC = \{Z,Z_1\}$, then we have $(ZZ_1;BC)=-1$. This means : $MB^2 = MZ \cdot MZ_1$ which implies that $M$ lies on the radical axis of $\{\omega_B,\odot(S,SZ)\}$ and similarly $\{\omega_C,\odot(S,SZ)\}$. Since the radical axis of these circles is the line through $M$ perpendicular to $\overline {O_BOO_C}$ , hence it is just $AM$. So the claim holds as $R\in AM$. Perform an inversion $\phi (R , \sqrt {RT \cdot RZ})$. Note that it fixes $\omega$ . Also we have $\phi (\Omega) = ZM \equiv BC$. Further note that since $\omega_B$ and $\omega_C$ are orthogonal to the circle of inversion, hence they are fixed by this inversion. Since $BC$ is tangent to $\omega_B$ and $\omega_C$ , we are done. $\blacksquare$

Insane problem but very cool! Solved with Rg230403, Pujnk Let $R$ be the midpoint of arc $BAC$, and let $RZ \cap \Gamma = T$. Let the tangent to $\Gamma$ at $T$ meet $BC$ at $Q$. Finally, let $Z'$ be the reflection of $Z$ across $Q$. Call the circle with center $Q$ and radius $QZ$ as $\Omega$. Finally, call the circle with center $K$ and radius $KP$ as $\gamma$ In $\triangle BTC$, since $TZ$ is the angle bisector (It passes through $R$) and $Q$ is the point on $BC$ with $QT$ tangent to $(BTC)$, its known that this means $QZ = QT$. So, $QZ^2 = QT^2 = QB.QC$, which means that $Q$ lies on $XY$ as well. Now let $\Omega$ meet $\Gamma$ for the second time at $J$. Since $\Omega$ is the $T$- apollonian circle in $\triangle BTC$, this means $\frac{BJ}{CJ} = \frac{BT}{CT} \implies JBTC$ is harmonic. So, $QJ$ is also tangent to $\Omega$ Since $\angle A'TZ = \angle ZTN = 90^\circ$ $Z', T, N$ are collinear and we see that $Z$ is the orthocenter of $\triangle RNZ'$. So, let $NZ$ and $RZ'$ meet at $J'$, which obviously has to lie on $\Gamma$. But since $\angle ZJZ' = 90^\circ$ as well, we see that $J'$ lies on $\Omega$ and $\Gamma$ and so $J' = J \implies R,J,Z'$ and $J,Z,N$ are collinear. Now, see that $\angle PNK = \angle NKM = \angle NZM = \angle JZZ' = \angle JTZ' = \angle KNZ' = \angle KNT$ (Where the last equality holds since $JT || KN || AM$), which means that $K$ is the midpoint of arc $PT$ in $\omega$ and so $KP = KT \implies T$ lies on $\gamma$ $\angle QTK = \angle QTZ + \angle ZTK = \angle QZT + \angle ZMK = \angle RZM + \angle ZMK$. But since $MK || NT$ and $NT \perp RT$, we get that $\angle RZM + \angle ZMK = 90^\circ \implies \angle QTK = 90^\circ$. So we see that $QT$ is tangent to $\gamma$ as well and so $\Gamma$ and $\gamma$ are tangent. Let $AP$ meet $TR$ at $S$. We will show that $S$ lies on $\gamma$ as well. This is because $\angle TSP = \angle JTR = \angle JNR = \angle ZNM = \angle ZTM = \angle STM$ and so $S$ is the reflection of $T$ across $MK$, since $K$ is the center of the circle, this means that $S$ lies on $\gamma$ as well. Now, by the homothety at $T$ taking $\Gamma$ to $\gamma$, we see that $S$ is the topmost point of $\gamma$ (Basically that the tangent to $\gamma$ at $S$ is parallel to $BC$). Now, let $\gamma '$ be the circle tangent to all of $\omega_b, \omega_c, \Gamma$ and let it meet $\omega_b$ at $U$, $\omega_c$ at $V$ and $\Gamma$ at $T'$ Since $Q$ is the exsimilicenter of $\omega_b, \omega_c$ and $U,V$ are the insimilicenters of $(\omega_b, \gamma'), (\omega_c, \gamma')$ respectively. So, by Monge on these three circles, we see that $U,V,Q$ are collinear. So, this means $BUVC$ is cyclic. Let $S'$ be the topmost point of $\gamma'$, then $S', U, B$ and $S', V, C$ are collinear. Since $SU.SB = SV.SC$, we see that $S$ lies on the radical axis of $\omega_b$ and $\omega_c$, which is just the median and so $S' \in AM$ Since $QB.QC = QU.QV$, $Q$ lies on the radical axis of $\gamma'$ and $\Gamma$ and so $QT$ is tangent to $\gamma'$. Since there is a unique circle passing through $S$ and $T$ tangent to $\Gamma$ at $T$, $\gamma' = \gamma$ and so we are done. $\blacksquare$

Very rich configuration. Let $\gamma$ be the circle with center $K$ and radius $KP$. Let $E$ be the center of $\omega_B$ and let $F$ be the center of $\omega_C$. Let $D=\overline{BC}\cap\overline{XY}$. Let $O$ denote the center of $\Gamma$. Claim: $\overline{AM}$ is the radical axis of $\omega_B,\omega_C$. Proof. Firstly, $M$ lies on the radical axis of $\omega_B,\omega_C$ as $MB^2=MC^2$. Let $R=\omega_B\cap \overline{AB}$ and $S=\omega_C\cap \overline{AC}$. I contend now that $BRSC$ is cyclic, from which it follows that $A$ lies on the radical axis of $\omega_B,\omega_C$ by radical axis theorem. It is sufficient to show that \begin{align*} AB\cdot AR&=AC\cdot AS\Longleftrightarrow\\ AB(AB-BR)&=AC(AC-CS)\Longleftrightarrow\\ AC^2-AB^2&=AM(BX-CY)\Longleftrightarrow\\ AC^2-AB^2&=2AM\cdot BC\cdot \cos{\angle AMB} \end{align*}where we used the fact that $\triangle AMB\sim \triangle BRX$ and $\triangle AMC\sim \triangle CYS$. By Law of Cosines on $\triangle AMB$ and $\triangle AMC$, we conclude desired. $\blacksquare$ [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,X,Y,M1,D,R,S,E,F; A=dir(110);B=dir(205);C=dir(335);M=midpoint(B--C);M1=intersectionpoints(A--100M-99A,circumcircle(A,B,C))[1];X=M1*A/B;Y=M1*A/C; D=extension(X,Y,C,B);E=intersectionpoint(perpendicular(B,line(B,C)),perpendicular(midpoint(B--X),line(B,X))); R=intersectionpoints(circle(E,abs(B-E)),A--B)[0];F=intersectionpoint(perpendicular(C,line(B,C)),perpendicular(midpoint(C--Y),line(C,Y))); S=intersectionpoints(circle(F,abs(C-F)),A--C)[0]; draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue);draw(circumcircle(B,R,S),royalblue+dashed);draw(circumcircle(R,B,X),royalblue);draw(circumcircle(S,Y,C),royalblue); draw(Y--D--B,heavyred+0.5); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy][/asy] Claim: Let $\overline{DQ}$ be tangent to $(ABC)$ so that $Q$ lies on the arc $BC$, not containing $A$. Then, $Q$ lies on $(ZMN)$. Proof. I contend that $\overline{XZ}$ passes through the midpoint of arc $BAC$, from which the shooting lemma finishes. We use ratio lemma, define $f(\bullet)=\pm\frac{\bullet B}{\bullet C}$, with choice of signs as usual. Hence, $f(Z)^2=f(X)f(Y)=f(D)=f(Q)^2$. We obtain desired. $\blacksquare$ [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,X,Y,M1,D,R,S,E,F,Q,Z,N; A=dir(110);B=dir(205);C=dir(335);M=midpoint(B--C);M1=intersectionpoints(A--100M-99A,circumcircle(A,B,C))[1];X=M1*A/B;Y=M1*A/C; D=extension(X,Y,C,B);E=intersectionpoint(perpendicular(B,line(B,C)),perpendicular(midpoint(B--X),line(B,X))); R=intersectionpoints(circle(E,abs(B-E)),A--B)[0];F=intersectionpoint(perpendicular(C,line(B,C)),perpendicular(midpoint(C--Y),line(C,Y))); S=intersectionpoints(circle(F,abs(C-F)),A--C)[0];Q=intersectionpoints(circumcircle(A,B,C),circumcircle(D,M,(0,0)))[1];Z=intersectionpoints(circle(D,abs(D-Q)),B--C)[0]; N=intersectionpoints(perpendicular(M,line(B,C)),circumcircle(A,B,C))[1]; draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue);draw(circumcircle(R,B,X),royalblue);draw(circumcircle(S,Y,C),royalblue); draw(Y--D--B,heavyred+0.5);draw(circumcircle(Z,N,M),royalblue);draw(D--Q,heavyred+0.5);draw(circumcircle(X,Y,Z),royalblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$Z$",Z,dir(Z)); dot("$N$",N,dir(N)); [/asy][/asy] Claim: $\gamma$ and $\Gamma$ are tangent to each other. Proof. Firstly, I claim that $K$ lies on $\overline{OQ}$. Indeed, $\measuredangle MQO=\measuredangle MDO=\measuredangle OMA=\measuredangle NMP=\measuredangle MQK$, where we used the facts that $ODMQ$ is cyclic and $\overline{AM}\perp \overline{EF}\equiv \overline{DO}$. Now $Q$ lies on $\gamma$ as $KQ=OQ-OK=OQ-\tfrac{OM\cdot ON}{OQ}=ON-OM=MN=KP$. We conclude that $\gamma$ and $\Gamma$ are tangent to each other. $\blacksquare$ [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,X,Y,M1,D,R,S,E,F,Q,Z,N,O,P,K; A=dir(110);B=dir(205);C=dir(335);M=midpoint(B--C);M1=intersectionpoints(A--100M-99A,circumcircle(A,B,C))[1];X=M1*A/B;Y=M1*A/C; D=extension(X,Y,C,B);E=intersectionpoint(perpendicular(B,line(B,C)),perpendicular(midpoint(B--X),line(B,X))); R=intersectionpoints(circle(E,abs(B-E)),A--B)[0];F=intersectionpoint(perpendicular(C,line(B,C)),perpendicular(midpoint(C--Y),line(C,Y))); S=intersectionpoints(circle(F,abs(C-F)),A--C)[0];Q=intersectionpoints(circumcircle(A,B,C),circumcircle(D,M,(0,0)))[1];Z=intersectionpoints(circle(D,abs(D-Q)),B--C)[0]; N=intersectionpoints(perpendicular(M,line(B,C)),circumcircle(A,B,C))[1];O=(0,0);P=intersectionpoints(A--M1,circumcircle(Z,M,N))[1];K=intersectionpoints(O--Q,circumcircle(Z,M,N))[0]; draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue);draw(circumcircle(R,B,X),royalblue);draw(circumcircle(S,Y,C),royalblue); draw(Y--D--B,heavyred+0.5);draw(circumcircle(Z,N,M),royalblue);draw(D--Q,heavyred+0.5);draw(D--F,heavyred+0.5);draw(Q--O--N,heavyred+0.5); draw(A--P,heavyred+0.5); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$Z$",Z,dir(Z)); dot("$N$",N,dir(N)); dot("$O$",O,dir(90)); dot("$P$",P,dir(P)); dot("$K$",K,dir(K)); [/asy][/asy] Claim: $\gamma$ is tangent to $\omega_B$ and $\omega_C$. Proof. Let $T=\omega_B\cap (PMB)$ and $U=\omega_C\cap (PMC)$. By radical axis thereom on $\omega_B,\omega_C, (PMB)$ and $\omega_B,(PMB),(PMC)$, we obtain that $\overline{BT},\overline{CS}$ concur on $\overline{AM}$ at, say, $L$. Thus, by PoP, $BTCS$ is cyclic quadrilateral. Also, $X,T,P$ are collinear as $\measuredangle XTB=\measuredangle XBD =\measuredangle AMB=\measuredangle PMB=\measuredangle PTB$. Similarly, $Y,U,P$ are collinear (yup that's right). Therefore, by PoP, $XTYU$ is cyclic quadrilateral. Now, radical axis on $(XTYU),(BTCU),(BCXY)$, we get that $D,T,U$ are collinear. Let $Q'=\overline{LZ}\cap (PMZ)$, note that by PoP, $Q'$ lies on $(BTZ)$ and $(CUZ)$. By inversion at $D$ with radius $\sqrt{DB\cdot DC}$, we get that $DQ'=DQ$ as inversion swaps $(BTZ)$ and $(CUZ)$. This is enough to conclude that $Q\equiv Q'$. Now, $\measuredangle UQT=\measuredangle UQZ+\measuredangle ZQT=\measuredangle CBL+\measuredangle LCB=\measuredangle ULT$ and $\measuredangle UPT=\measuredangle UPM+\measuredangle MPT=\measuredangle CBL+\measuredangle LBC=\measuredangle ULT$, hence $QPULC$ is cyclic. Let $K'$ be the center of $(QPULC)$. We have $K', K, P$ collinear since $\measuredangle MPK=\measuredangle NMP=90^\circ+\measuredangle BMA=90^\circ+ \measuredangle DBX=\measuredangle EBX=90^\circ+\measuredangle BTX=90^\circ+\measuredangle LTP=\measuredangle MPK'$. As both $K,K'$ lie on the perpendicular bisector of $\overline{PQ}$, this all is enough to conclude that $K\equiv K'$. As $\triangle TBX$ and $\triangle TLP$ are homothetic, we conclude that $\gamma$ and $\omega_B$ are tangent to each other. Similarly, $\gamma$ and $\omega_C$ are tangent to each other. $\blacksquare$ [asy][asy] import olympiad;import geometry; size(12cm);defaultpen(fontsize(10pt)); pair A,B,C,M,X,Y,M1,D,R,S,E,F,Q,Z,N,O,P,K,T,U,L; A=dir(110);B=dir(205);C=dir(335);M=midpoint(B--C);M1=intersectionpoints(A--100M-99A,circumcircle(A,B,C))[1];X=M1*A/B;Y=M1*A/C; D=extension(X,Y,C,B);E=intersectionpoint(perpendicular(B,line(B,C)),perpendicular(midpoint(B--X),line(B,X))); R=intersectionpoints(circle(E,abs(B-E)),A--B)[0];F=intersectionpoint(perpendicular(C,line(B,C)),perpendicular(midpoint(C--Y),line(C,Y))); S=intersectionpoints(circle(F,abs(C-F)),A--C)[0];Q=intersectionpoints(circumcircle(A,B,C),circumcircle(D,M,(0,0)))[1];Z=intersectionpoints(circle(D,abs(D-Q)),B--C)[0]; N=intersectionpoints(perpendicular(M,line(B,C)),circumcircle(A,B,C))[1];O=(0,0);P=intersectionpoints(A--M1,circumcircle(Z,M,N))[1];K=intersectionpoints(O--Q,circumcircle(Z,M,N))[0]; T=intersectionpoints(circle(E,abs(B-E)),circumcircle(P,M,B))[1];U=intersectionpoints(circle(F,abs(C-F)),circumcircle(P,M,C))[0];L=extension(B,T,C,U); draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue);draw(circumcircle(R,B,X),royalblue);draw(circumcircle(S,Y,C),royalblue); draw(Y--D--B,heavyred+0.5);draw(circumcircle(Z,N,M),royalblue);draw(D--Q,heavyred+0.5);draw(D--F,heavyred+0.5); draw(A--P,heavyred+0.5);draw(circumcircle(B,P,M),lightblue+0.5);draw(circumcircle(C,P,M),lightblue+0.5); draw(B--L--C,lightmagenta+0.5); draw(X--P--Y,magenta+dotted);draw(D--U,heavyred+0.5+dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$Z$",Z,dir(Z)); dot("$N$",N,dir(N)); dot("$O$",O,dir(90)); dot("$P$",P,dir(P)); dot("$K$",K,dir(K)); dot("$T$",T,dir(T)); dot("$U$",U,dir(U)); dot("$L$",L,dir(L)); [/asy][/asy]

Next we present a very quick way to show $\Omega$ is tangent to $\omega_b$ (which clearly completes the proof). We use the same diagram. [asy][asy] import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,X,Y,M1,D,R,S,E,F,Q,Z,N,O,P,K; A=dir(110);B=dir(205);C=dir(335);M=midpoint(B--C);M1=intersectionpoints(A--100M-99A,circumcircle(A,B,C))[1];X=M1*A/B;Y=M1*A/C; D=extension(X,Y,C,B);E=intersectionpoint(perpendicular(B,line(B,C)),perpendicular(midpoint(B--X),line(B,X))); R=intersectionpoints(circle(E,abs(B-E)),A--B)[0];F=intersectionpoint(perpendicular(C,line(B,C)),perpendicular(midpoint(C--Y),line(C,Y))); S=intersectionpoints(circle(F,abs(C-F)),A--C)[0];Q=intersectionpoints(circumcircle(A,B,C),circumcircle(D,M,(0,0)))[1];Z=intersectionpoints(circle(D,abs(D-Q)),B--C)[0]; N=intersectionpoints(perpendicular(M,line(B,C)),circumcircle(A,B,C))[1];O=(0,0);P=intersectionpoints(A--M1,circumcircle(Z,M,N))[1];K=intersectionpoints(O--Q,circumcircle(Z,M,N))[0]; draw(A--B--C--cycle,heavyred+0.5); draw(circumcircle(A,B,C),royalblue);draw(circumcircle(R,B,X),royalblue);draw(circumcircle(S,Y,C),royalblue); draw(Y--D--B,heavyred+0.5);draw(circumcircle(Z,N,M),royalblue);draw(D--Q,heavyred+0.5);draw(D--F,heavyred+0.5);draw(Q--O--N,heavyred+0.5); draw(A--P,heavyred+0.5); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$Z$",Z,dir(Z)); dot("$N$",N,dir(N)); dot("$O$",O,dir(90)); dot("$P$",P,dir(P)); dot("$K$",K,dir(K)); [/asy][/asy] We desire $EK = EB + MN$. Let $W$ be projection of $N$ onto $EB$. Clearly $BMNW$ is rectangle, so $$ EB + MN = EB + BW = EW $$Let $\gamma_1 = \odot(O,OM)$ and $\gamma_2 = \odot(E,EW)$. We want $K \in \gamma_2$. We have already show $K \in \gamma_1$ (since $\Gamma$ is tangent to $\Omega$). Since radical axes of $\gamma_1,\gamma_2$ is parallel to $NK$, so it suffices to show $$ N \text{ lies on radical axes of } \gamma_1, \gamma_1 $$But observe \begin{align*} \text{Pow}(N,\gamma_1) = ON^2 - OM^2 = OB^2 - OM^2 = BM^2 = NW^2 = \text{Pow}(N,\gamma_2) \end{align*}We are done! $\blacksquare$