APMO

Let $N$ be the radical center of the circumcircles of triangles $ABC$, $BMP$ and $CMP$. The pairwise radical axes of these circles are $BD$, $CE$ and $PM$, and hence they concur at $N$. Now, note that in directed angles: $$\angle MCE=\angle MPE=\angle MPY=\angle MBY.$$ It follows that $BY$ is parallel to $CE$, and analogously that $CX$ is parallel to $BD$. Then, if $L$ is the intersection of $BY$ and $CX$, it follows that $BNCL$ is a parallelogram. Since $BM=MC$ we deduce that $L$ is the reflection of $N$ with respect to $M$, and therefore $L\in AM$. Using power of a point from $L$ to the circumcircles of triangles $BPM$ and $CPM$, we have $$LY\cdot LB=LP\cdot LM=LX\cdot LC.$$ Hence, $BYXC$ is cyclic. Using the cyclic quadrilateral we find in directed angles: $$\angle LXY=\angle LBC=\angle BCN=\angle NDE.$$ Since $CX\parallel BN$, it follows that $XY\parallel DE$. Let $Q$ and $R$ be two points in $\Gamma$ such that $CQ$, $BR$, and $AM$ are all parallel. Then in directed angles: $$\angle QDB=\angle QCB=\angle AMB=\angle PMB=\angle PDB.$$ Then $D$, $P$, $Q$ are collinear. Analogously $E$, $P$, $R$ are collinear. From here we get $\angle PRQ=\angle PDE=\angle PXY$, since $XY$ and $DE$ are parallel. Therefore $QRYX$ is cyclic. Let $S$ be the radical center of the circumcircle of triangle $ABC$ and the circles $BCYX$ and $QRYX$. This point lies in the lines $BC$, $QR$ and $XY$ because these are the radical axes of the circles. Let $T$ be the second intersection of $AS$ with $\Gamma$. By power of a point from $S$ to the circumcircle of $ABC$ and the circle $BCXY$ we have $$SX\cdot SY=SB\cdot SC=ST\cdot SA.$$ Therefore $T$ is in the circumcircle of triangle $AXY$. Since $Q$ and $R$ are fixed regardless of the choice of $P$, then $S$ is also fixed, since it is the intersection of $QR$ and $BC$. This implies $T$ is also fixed, and therefore, the circumcircle of triangle $AXY$ goes through $T\ne A$ for any choice of $P$.

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Alternative solution.

Let the lines $DP$ and $EP$ meet the circumcircle of $ABC$ again at $Q$ and $R$, respectively. Then $\angle DQC=\angle DBC=\angle DPM$, so $QC\parallel PM$. Similarly, $RB\parallel PM$. Now, $\angle QCB=\angle PMB=\angle PXC=\angle (QX,CX)$, which is half of the arc $QC$ in the circumcircle ${\omega }_C$ of $QXC$. So ${\omega }_C$ is tangent to $BS$; analogously, ${\omega }_B$, the circumcircle of $RYB$, is also tangent to $BC$. Since $BR\parallel CQ$, the inscribed trapezoid $BRQC$ is isosceles, and by symmetry $QR$ is also tangent to both circles, and the common perpendicular bisector of $BR$ and $CQ$ passes through the centers of ${\omega }_B$ and ${\omega }_C$. Since $MB=MC$ and $PM\parallel BR\parallel CQ$, the line $PM$ is the radical axis of ${\omega }_B$ and ${\omega }_C$. However, $PM$ is also the radical axis of the circumcircles ${\gamma }_B$ of $PMB$ and ${\gamma }_C$ of $PMC$. Let $CX$ and $PM$ meet at $Z$. Let $p(K,\omega )$ denote the power of a point $K$ with respect to a circumference $\omega$. We have $$p\left(Z,{\gamma }_B\right)=p\left(Z,{\gamma }_C\right)=ZX\cdot ZC=p\left(Z,{\omega }_B\right)=p\left(Z,{\omega }_C\right).$$ Point $Z$ is thus the radical center of ${\gamma }_B$, ${\gamma }_C$, ${\omega }_B$, ${\omega }_C$. Thus, the radical axes $BY$, $CX$, $PM$ meet at $Z$. From here, $$\begin{array}{rl}& ZY\cdot ZB=ZC\cdot ZX\Rightarrow BCXY\text{ cyclic}\\ & PY\cdot PR=PX\cdot PQ\Rightarrow QRXT\text{ cyclic.}\end{array}$$ We may now finish as in Solution 1. $\square$