Hint: Let ray $HM \cap (ABC) = A'$ Then $A'$ lies on $(PQN)$ and $A'$ is center of $(AB'C')$. In some moment I remembered idea in problem 10 from Correspondence round and after this I solved this problem...

Nice and not easy problem, i will just post what i needed to solve. Let HM intersects (ABC) at point T, then HCBT is parallelogram, then HPT is isosceles.Firstly note that AT is diameter of (ABC), which gives us by homotety (with coefficent 2) that T is center of (AB'C'). Observe that Q is orthocenter of HPN, then using this you can obtain that (PQTN) is cyclic. We know that Radiues of circles (ABC), (BHC), (PQN) are equal (denote their radius R), and we know that radius of AB'C' is 2R, using that (PQN) passes through T, which center of AB'C', and using that radius of AB'C' is 2 times of (PQN) we obtain that the circles (PQN) and (AB'C') are tangent to each other.

Here goes another sketch of the insane day 2

Solved with $\textbf{egxa}$ Let $HM$ intersects arc $BC$ of $\Gamma$ not containing $A$ at the point $D$, and arc $BC$ of $\Gamma$ containing $A$ at the point $E$. Let intersection of $HQ$ and $\Gamma$ be $R \neq Q$. Consider homothety with center $M$ and ratio $-1$. Then $H,B,C$ will be the images of $D,C,B$ respectively. Thus this homothety sends $\Gamma$ to $\omega$. Since the further point goes as far as the further point with this homothety, we conclude that this homothety sends $R$ to $P$. Thus we obtain $MR=MP$. Furthermore with $PoP$ $$MD \cdot MN=MH \cdot MN=MB \cdot MC=MQ \cdot MR=MQ \cdot MP$$Thus points $P, Q, D, N$ are concyclic. And also from the homothety we have mentioned before, we obtain that reflaction of $A$ about $M$ is on $\omega$. (Let $M'$ be that point.) This point is also the midpoint of $B'C'$. Also from homothety with center $A$ and ratio $2$ we obtain that $D$ is the center of $(AB'C')$ (Since $D$ is the antipode of $A$.) Let the intersection of $M'N$ and $(PQDN)$ be $K \neq N$. From the first homothety we have mentioned, we obtain $AN \perp NK$ since $AE \perp EM$. Thus $DK$ is the diameter of $(PQDNK)$. Also since $AD$ is the diameter of $\Gamma$ and $DK$ is the diameter of $(PQDNK)$ we obtain $AQ \perp QD$ and $QK \perp QD$ . Thus points $A,Q,K$ are collinear. And from $KM'||QM$ and $AM=MM'$ we obtain $AQ=QK$ and $AD=DK$. Since $D$ is the center of $(AB'C')$ we obtain points $A, B', K, C'$ are concylic. Thus $K$ is on both of the circles $(AB'C')$ and $(PQN)$. Furthermore the line which contains the center of $(AB'C')$ and the center of $(PQN)$ passes through the point $K$ since $DK$ is the diameter of $(PQN)$. Thus we obtain that these both circles are tangent at the point $K$. $Q.E.D.$

Let $HM\cap \odot(ABC)=T$ Note that $\angle BCN=\angle BHT=\angle HTC$ So $MC^{2}=MT\cdot MN$ Similarly $MB^{2}=MQ\cdot MP$ Which means that $Q,P,N,T$ are cyclic Consider that $AT$ is the diameter of $\odot(ABC)$ Hence $T$ is the circumcenter of $\odot(AB'C')$ Suppose the radius of $\odot(ABC)$ and $\odot(PQN)$ is $R$ and $r$,respectively So we just need to prove that $R=2r$ Note that $\angle QPT=\angle MTQ=\angle THQ$ Hence $MQ\bot BN$ Notice $PQ\bot HT$ Hence $Q$ is the orthocenter of $\triangle HPN$ Hence $\angle PQN=180^{\circ}-\angle PHN$ Which means that $\odot(PQN)$ and $\odot(BHC)$ have equal radius Since $\odot(BHC)$ and $\odot(ABC)$ have equal radius Hence $\odot(PQN)$ and $\odot(ABC)$ have equal radius So $R=2r$

Kind of oversaw the main trick leading into a sad 30 minutes solve. Using $(BC)$ inversion we get that if we let $A'$ the A-antipode of $A$ in $\Gamma$ then $PQA'N$ are cyclic (becuase it sends $\Gamma$ to $\omega$), now by homothety $A'$ is the center of $(AB'C')$ and also let $Q'$ the reflection of $Q$ over $M$ then clearly $Q'$ lies in $\Gamma$ and $Q'M \cdot MP=HM \cdot MN$ hence $Q$ is ortocenter of $\triangle HNP$ meaning that $(NPQ)$ and $\omega$ have the same radius so $(NPQ)$ and $\Gamma$ have the same radius, now let $M_A$ the midpoint of $B'C'$, then clearly $\angle HNM_A=90$ hence we have that $AQ \cap NM_A=T$ lies in $(NPQ)$ and also $AQ=QT$ holds by pitagoras, so by Homothety of scale 2 $AB'TC'$ is cyclic, now since $\angle A'NT=90$ we get $(PNQ), (AB'C')$ tangent at $T$ as desired, thus we are done

$(HBC)$ is symmetric to $(ABC)$ with center $M$ So we get that $MA'*MN=MQ*MP$ which gives that $Q$ is the orthocenter of $HPN$ and that $(PQA'N)$ are cyclic. Let $T=NM_A\cap AQ$ $\angle TQA'=180-\angle AQA'=90=\angle HDM_A=\angle HNA$ So $T$ belongs to $(PQA'N)$. $NM_A//QM$ so $Q$ is the midpoint of $AT$ but $A'$ is the center of $AB'C'$ and $\angle A'QA=90$ so $T$ belongs to $(AB'C')$. Now since $(ABC),(PQN),(HBC)$ have the same radious say $r$ and the radioys of $(AB'C')$ is $2r$ with $\angle TQA'=90$ we have the result