Construct the foot of perpendicular $D$ from $A$ to $X_2Y_1$. Since $\angle AYY_1 = \angle ABY_1 = \angle AXX_2 = \angle ACX_2 = 90^\circ$, the five points $DX_2XAC$ lie on a common circle, and likewise the five points $DY_1YAB$ lie on a common circle. Considering the radical center of these two circles together with the originally given circle (three circles in total), we obtain that the three lines $AD$, $BY$, $CX$ are concurrent. Let this common point be $S$, and we have
$$SA \cdot SD = SB \cdot SY = SC \cdot SX.$$
Consider the transformation centered at $S$ with inversive power $SA \cdot SD$. This transformation interchanges $AD$, $BY$, $CX$ pairwise; and since $BAC$ are collinear with $A$ the midpoint of $BC$, it follows that $DXSY$ lie on a common circle and form a harmonic quadrilateral. We have
$$\angle DZX = \angle DYX + \angle ZDY = \angle DSX + \angle XDS = 180^\circ - \angle DXS \\ \angle SZX = \angle SYX + \angle ZSY = \angle SDX + \angle XSD = 180^\circ - \angle DXS.$$
(Here we used the fact that $DXSY$ is harmonic, so $DS$, $DZ$ are isogonal conjugates with respect to $\angle XDY$, and $SD$, $SZ$ are isogonal conjugates with respect to $\angle XSY$.) Hence $\angle DZX = \angle SZX$. Also, since $XYBC$ lie on a common circle, $\triangle SXY$ is similar to $\triangle SBC$, so $\angle SAB = \angle SZX = \angle DZX$. Let $XY$ meet $BC$ at point $T$; then the four points $DZAT$ lie on a common circle.
Let the center of the originally given circle be $O$. Since $OZ \perp ZT$ and $OA \perp AT$, the five points $DZOAT$ lie on a common circle. Let $L_3$ be the line through $O$ parallel to $BC$, and let $L_4$ be the line through $Z$ perpendicular to $AZ$. Then the antipodal point of $A$ on the circle $DZOAT$ is the common intersection point of $X_2Y_1$, $L_3$, $L_4$, that is, $X_2Y_1$, $L_3$, $L_4$ are concurrent. By the same argument, $X_1Y_2$, $L_3$, $L_4$ are concurrent. Hence this point is exactly $P$, and therefore $\angle AZP = 90^\circ$. This completes the proof.