Redefine things, let $Q'$ be the point such that $\measuredangle Q'XY=\measuredangle ACP$ and $\measuredangle Q'YX=\measuredangle ABP$ and let $Q$ be the point such that $\measuredangle QXY=\measuredangle ABP$ and $\measuredangle QYX=\measuredangle ACP$. Note that $XYQQ'$ is isosceles trapezoid, hence we do not care about $Q'$ anymore and care about $Q$ only. I contend that the desired fixed point is the intersection of $(ABC)$ and $A$-symmedian of $\triangle ABC$. Claim: $(PXB),(PYC),(ABC)$ and $(XYQ)$ concur at single point. Proof. Let $D=(PXB)\cap (PYC)$. Indeed, \begin{align*} \measuredangle CDB&=\measuredangle CDP+\measuredangle PDB=\measuredangle CYP+\measuredangle PXB\\&=\measuredangle YXA+\measuredangle AYX=\measuredangle YAX=\measuredangle CAB \end{align*}and \begin{align*} \measuredangle YDX=\measuredangle YDP+\measuredangle PDX=\measuredangle QYX+\measuredangle YXQ=\measuredangle YQX.\,\square \end{align*} Let $E=(ABC)\cap (XYQ)$. Claim: $A,Q,E$ are collinear. Proof. Indeed, \begin{align*} \measuredangle QED=\measuredangle QXD=\measuredangle XBD=\measuredangle ABD=\measuredangle AED.\,\square \end{align*}Hence, we now would like to show that $AQ$ is a fixed line. We use sine bash. Claim: $\frac{PX}{PY}=\frac{AC\cdot BX}{AB\cdot CY}$. Proof. Since $ZX=ZY$, we have $\measuredangle XTB=\measuredangle XYZ=\measuredangle ZXY=\measuredangle ZTY=\measuredangle CTY\implies PT$ bisects $\angle XTY$. Hence, \begin{align*} \frac{PX}{PY}=\frac{XT}{TY}=\frac{\frac{XB\sin{\angle ABC}}{\sin{\angle XTB}}}{\frac{CY\sin{\angle ACB}}{\sin{\angle CTY}}}=\frac{BX}{CY}\cdot \frac{\sin{\angle ABC}}{\sin{\angle ACB}}=\frac{BX}{CY}\cdot \frac{AC}{AB}.\,\square \end{align*} Now, \begin{align*} \frac{\sin{XAQ}}{\sin{QAY}}&=\frac{\frac{XQ\sin{\angle AXQ}}{AQ}}{\frac{YQ\sin{\angle AYQ}}{AQ}}=\frac{XQ\sin{\angle AXQ}}{YQ\sin{\angle AYQ}}=\frac{\sin{\angle XYQ}\sin{\angle AXQ}}{\sin{\angle YXQ}\sin{\angle AYQ}}\\&=\frac{\sin{\angle PCY}\sin{\angle XPB}}{\sin{\angle PBX}\sin{\angle CPY}}=\frac{PY}{CY}\cdot \frac{BX}{PX}=\frac{AB}{AC}. \end{align*}We are done. $\blacksquare$ [asy][asy] import geometry; size(12cm);defaultpen(fontsize(10pt)); pair A,B,C,X,Y,Z,T,P,Q,D,E; A=dir(120); B=dir(200); C=dir(340); X=0.55A+0.45B;Y=0.75C+0.25A;Z=intersectionpoint(line(B,C),perpendicular(midpoint(X--Y),line(X,Y))); T=intersectionpoints(circumcircle(X,Y,Z),B--C)[1];P=intersectionpoint(perpendicular(T,line(B,C)),line(X,Y)); D=intersectionpoints(circumcircle(P,X,B),circumcircle(A,B,C))[1];E=intersectionpoints(circumcircle(X,Y,D),circumcircle(A,B,C))[1]; Q=intersectionpoints(A--E,circumcircle(D,X,Y))[0]; draw(A--B--C--cycle, red+1);draw(circumcircle(A,B,C),royalblue); draw(circumcircle(P,X,B),cyan);draw(circumcircle(P,Y,C),cyan);draw(circumcircle(X,Y,D),royalblue);draw(A--E,red+1); draw(X--Y,red+1);draw(X--Q,orange+0.5);draw(Y--Q,orange+0.5);draw(circumcircle(X,Y,Z),royalblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$T$",T,dir(T)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$Q$",Q,dir(Q)); [/asy][/asy]

here is solution using harmonic division: let $BP,CP$ intersect $(ABC)$ at $E,F$. then by pascal's theorem $EY,FX$ intersect at $S$ which lies on $(ABC)$. by simple angle chasing we get that it is also on $(XYQ)$. let $CX\cap (ABC)=F'$. let $XY\cap BC=G$. then because $\widehat{PTZ}=90$ and because $TZ$ is the external bisector of $\widehat{XTZ}$ then $PT$ is the internal bisector. thus $(GP,XY)=-1$ so $BY,CX,AP$ are concurrent. call this point $R$. also let $AP \cap BC=D$. by looking through point C, we get that $(DP,RA)=(GP,XY)=-1$. now: $$(BC,SA)=(AF',FB)=(AR,PD)=-1$$where the first one is projecting through $X$ onto $(ABC)$ and the second equality is by procting $(ABC)$ onto $AP$ through $C$. so we get that $S$ is fixed.(the intersection of $(ABC)$ and the $A$-symmedian.) we are done

Since $\angle AXY + \angle AYX = \angle 180 - \angle A$ so $BXP$ and $CYP$ and $ABC$ meet in a point. Let the point be $S$. Claim $: XQYS$ is cyclic. Proof $:$ Note that $\angle QXS + \angle QYS = \angle QXY + \angle PXS + \angle QYX + \angle PYX = \angle ACP + \angle PBS + \angle ABP + \angle PCS = \angle 180$. It's Easy to notice that with moving $Y$ and fixing $X$, $S$ moves on $ABC$ so $S$ is not our desired point. Let $XQY$ meet $ABC$ for the second time at $S'$. This must be our fixed point. Claim $: AS'$ passes through reflection of $Q$ about perpendicular bisector of $XY$. Let this reflected point be $Q'$. Note that since $XQYS'S$ is cyclic then $Q'$ lies on this circle as well. $\angle Q'S'S = \angle Q'YS = \angle Q'YX + \angle XYS = \angle QXY + \angle PCS = \angle ACS = \angle AS'S$ so $A,Q',S'$ are collinear. We need to show that $AQ'$ is a fixed line. Let $AQ'$ meet $BC$ at $K$. we need to show that $K$ is fixed. $\frac{BK}{KC} = \frac{AB}{AC}.\frac{\sin{Q'AB}}{\sin{Q'AC}}$ so we need to prove $\frac{\sin{Q'AB}}{\sin{Q'AC}}$ is fixed. Claim $: \frac{XQ'}{YQ'} = \frac{\sin{Q'AX}}{\sin{Q'AY}} . \frac{\sin{AYQ'}}{\sin{AXQ'}}$. Proof $:$ $\frac{XQ'}{YQ'} = \frac{\frac{XQ'}{Q'A}}{\frac{YQ'}{Q'A}} = \frac{\sin{Q'AX}}{\sin{Q'AY}} . \frac{\sin{AYQ'}}{\sin{AXQ'}}$. So we have $\frac{\sin{Q'AB}}{\sin{Q'AC}} = \frac{XQ'}{YQ'} . \frac{\sin{AXQ'}}{\sin{AYQ'}}$. Claim $: Q,P,S$ are collinear. Proof $:$ Note that $\angle XSP = \angle XBP = \angle ABP = \angle QYX = \angle QSX$. Claim $: Q'Y$ and $Q'X$ are tangent to $PYC$ and $PXB$. Proof $:$ Note that $\angle Q'YX = \angle QXY = \angle ACP = \angle YCP$ and $\angle Q'XY = \angle QYX = \angle ABP = \angle XBP$. Now we have $\frac{XQ'}{YQ'} . \frac{\sin{AXQ'}}{\sin{AYQ'}} = \frac{YQ}{XQ} . \frac{\sin{AXQ'}}{\sin{AYQ'}} = \frac{\sin{YSP}}{\sin{XSP}} . \frac{\sin{XPB}}{\sin{YPC}} = \frac{\sin{YCP}}{\sin{XBP}} . \frac{\sin{XPB}}{\sin{YPC}} = \frac{YP}{YC} . \frac{XB}{XP}$. $BX = \frac{XT.\sin{XTB}}{\sin{B}}$ and $CY = \frac{TY.\sin{YTC}}{\sin{C}}$ and $\frac{YP}{XP} = \frac{YT}{XT} . \frac{\sin{YTP}}{\sin{XTP}}$ so $\frac{YP}{YC} . \frac{XB}{XP} = \frac{XT.\sin{XTB}}{\sin{B}} . \frac{\sin{C}}{TY.\sin{YTC}} . \frac{YT}{XT} . \frac{\sin{YTP}}{\sin{XTP}} = \frac{\sin{XTB}}{\sin{YTC}} . \frac{\sin{YTP}}{\sin{XTP}} . \frac{\sin{C}}{\sin{B}}$ Since $\angle XTB = \angle XYZ = \angle YXZ = \angle YTC$ and $\angle YTP = \angle 90 - \angle YTC = \angle 90 - \angle XTB = \angle PTX$ then $\frac{\sin{XTB}}{\sin{YTC}} . \frac{\sin{YTP}}{\sin{XTP}} . \frac{\sin{C}}{\sin{B}} = \frac{\sin{C}}{\sin{B}}$ so $\frac{\sin{Q'AB}}{\sin{Q'AC}} = \frac{\sin{C}}{\sin{B}}$ is fixed as wanted. we're Done.

During the solution i found a interesting lemma that may be useful for life: Lemma: Let $\triangle ABC$ be a triangle and $D,E,F$ points on $BC, CA, AB$ such that $AD, BE$ and $CF$ concur. Let $M$ be the intersection of $(AEF), (BDF)$ and $(CDE)$. Let $N$ be the second intersection of $(MBC)$ with $(AEF)$. Then, $(A,N;E,F) = -1$. Proof. Consider a inversion wrt $M$ and arbitrary radius. Using the cyclic quadrilaterals and collinearities, the inverted figure will be points $A',B',C'$ on sides $E'F', D'F', D'E'$ of triangle $D'E'F'$ and $N' = B'C' \cap E'F'$. Note that $D'A', E'B', F'C'$ concur by Menelaus Theorem, since $$\dfrac{E'A' \cdot F'B' \cdot D'C'}{A'F' \cdot B'D' \cdot C'E'} = \dfrac{EA \cdot FB \cdot CD}{AF \cdot BD \cdot CE} = -1 $$ where we're using the Inversion Distance Formula and the concurrence of the statement. Hence, since $ D'A', E'B', F'C'$ concur, $(A',N';E',F') = -1 \Rightarrow (A,N;E,F) = -1$, as Inversion preserves cross ratio. $\blacksquare$ Now let's attack the problem. Claim 1: $(PXB), (PYC), (ABC)$ and $(XYQ)$ concur. Proof. Let $D=(PXB) \cap PYC$. We have $\angle AXY = \angle PDB$ and $\angle AYX = \angle PDC$, so $\angle BAC + \angle BDC=\angle BAC + \angle AXY + \angle AYX = 180^\circ \Rightarrow D \in (ABC)$. Now, $\angle PDX = \angle ABP = \angle QYX$ and $\angle PDY = \angle PCY = \angle QXY$, hence $\angle XQY + \angle XDY = 180^\circ \Rightarrow D \in (XYQ)$. $\blacksquare$ Claim 2: $AP, BY$ and $CX$ concur. Proof. Let $XY \cap BC = E$. Since $Z$ is the mid point of arc $XY$ and $\angle ZTP = 90 ^\circ$, we have that $TP$ bisects $\angle XTY$ and then $(X,Y;P,E) = -1$. But if $BY \cap CX = F$, $AF \cap BC = G$, it holds that $(B,C;G,E) = -1$ and projecting by A we have $P,T,F$ collinears. $\blacksquare$ Now, by the presented Lemma in triangle $\triangle AXY$ and points $P,C,B$, if $(DXY) \cap (ABC) = S$, then $(B,C;A,S) = -1 \Rightarrow$ $(QXY)$ pass through a fixed point (the point $S$). $\blacksquare$

Solved with Om245.High quality problem. Not too hard once you guess the fixed point though. We claim that as $X$ and $Y$ move along $AB$ and $AC$, the circle $(QXY)$ always passes through the point which is the intersection of the $A-$symmedian and $(ABC)$. We start off with some preliminary observations which will come in useful later. Claim :Lines $\overline{TP}$ and $\overline{BC}$ are respectively the internal and external $\angle XTY$-bisectors. Proof : Note that $Z$ is the major arc midpoint of $XY$ in $(XTY)$ by the nature of its definition. Thus, $TZ$ is the external $\angle XTY$-bisector and $\overline{TZ}$ is simply $\overline{BC}$. Further, since $TP \perp TZ$, it also follows that $\overline{TP}$ is the internal $\angle XTY$-bisector which finishes the proof of the claim. Now, let $K = (BXP) \cap (CYP)$ and let $R = (XKY) \cap (ABC)$. We now locate $K$. Claim : The point $K$ lies on both circles $(XQY)$ and $(ABC)$. Proof : This is a simple angle chase. Note that, \[\measuredangle BKC = \measuredangle BKP + \measuredangle PKC = \measuredangle AXY + \measuredangle AYX = \measuredangle XAY = \measuredangle BAC \]which implies that $K$ lies on $(ABC)$. Almost similarly, \[\measuredangle XKY = \measuredangle XKP + \measuredangle PXY = \measuredangle XBP + \measuredangle PCY = \measuredangle QXY + \measuredangle XYQ = \measuredangle XQY\]which implies that $K$ also lies on $(XQY)$ as claimed. Claim : Lines $AP$, $BY$ and $CX$ concur. Proof : For this, we simply use Ceva's on $\triangle AXY$. We wish to have, \[\frac{AB}{BX} \cdot \frac{XP}{PY}\cdot \frac{CY}{AC}=1\]Note that this rewrites to, \[\frac{AB}{AC} = \frac{BX}{XP} \cdot \frac{YP}{YC} = \frac{BX}{CY} \cdot \frac{YT}{XT}\]due to our previous observation that $PT$ is the internal $\angle XTY$-bisector. Now, we use the sine rule to see that we simply require, \[\frac{\sin C}{\sin B} = \frac{\sin BTX}{\sin B} \cdot \frac{\sin C}{\sin CTY}\]which is immediately true due to our previous result that $BC$ is the external $\angle XTY$-bisector (so $\measuredangle BTX = \measuredangle YTC$), which finishes the proof of the reduced problem. Thus, the claim is proved. We perform an inversion around $K$ with arbitrary radius, and we let $\mathcal{P}^*$ denote image of $\mathcal{P}$ under this inversion. First, note that points $B^*$ , $X^*$ and $P^*$ and $C^*$ , $Y^*$ and $P^*$ are collinear. Now from our previous claim that $AP$ , $BY$ and $CX$ concur, \[\frac{AB}{BX} \cdot \frac{XP}{PY}\cdot \frac{CY}{AC}=1\]we get that (since inversion preserves ratios) \[\frac{A^*B^*}{A^*C^*} \cdot \frac{C^*Y^*}{Y^*P^*}\cdot \frac{P^*X^*}{X^*B^*}=1\]By the Ceva/Menaluas picture we then have $(A^*,R^*;B^*,C^*)=-1$ (remember after inversion, $R^* = X^*Y^* \cap B^*C^*$). Returning to the original diagram, we have that $(A,R;B,C)=-1$ we get that $R$ is in fact the intersection of the $A-$symmedian with $(ABC)$ as desired. Hence $(QXY)$ passes through a fixed point as $X$ and $Y$ vary as claimed.

The IGO Grind begins!. Let $Z'$ the antipode of $Z$ in $(XYZ)$, trivially $Z',P,T$ are collinear, so if we let $BC \cap XY=G$ then by external angle bisector we have that $-1=(G, P; X, Y)$ so now let $BP,CP$ hit $(ABC)$ at $E,F$ respectively, note that from converse of Pascal we have that $EY \cap FX=K$ lies on $(ABC)$, now let $GK \cap (ABC)=J$ and let $JX,JY$ hit $(ABC)$ at $U,V$. Now from DDIT on $BXYC$ we have that $(KB, KC), (KX, KY), (KG, KA)$ are pairs of involution so projecting onto $(ABC)$ gives $A,P,J$ collinear, and projecting from $J$ gives $-1=(A, K; U, V)$. But also from DDIT at $BXYC$ we have that $(JB, JC), (JX, JY), (JA, JG)$ are pairs of involution so projecting onto $(ABC)$ gives $BV, CU, AK$ are concurrent, projecting from this concurrency point we have that $-1=(A, K; B, C)$ so $AK$ is symedian. $$\angle XQY=180-\angle ABP-\angle ACP=180-\angle AKE-\angle AKF=180-\angle XKY$$Which implies $(XQYK)$ cyclic so our fixed point is $K$ thus we are done .