Let $P_1$ be the point where the circumcircle of $ACY$ intersects $BC$ for the second time. Let $D$ be the foot of the bisector of $\angle BAC$. Clearly, $D$ is interior to the circumcircle of $ACY$ and by power of a point, we have that
$$DP_1 \cdot DC = DA \cdot DY.$$
Since $BX$ and $CY$ are parallel, $\frac{DB}{DC} = \frac{DX}{DY}$, from where it is easily obtained that $DP_1 \cdot DB = DA \cdot DX$ and as $D$ is external to segments $BP_1$ and $AX$, $AXP_1B$ is cyclic. For the two possible positions (see the diagrams below), using the cyclic quadrilaterals $ACYP_1$ and $AXP_1B$ we have that $\angle CP_1Z = \angle DP_1X = \angle BAX = \angle CAY = \angle CP_1Y$. Then $P_1$, $Z$ and $Y$ are collinear, and $P_1$ is the point where $YZ$ intersects $BC$, from where $P_1 = P$ and $AXBP$ and $ACYP$ are cyclic. Now let $S$ and $T$ be the circumcenters of $ABX$ and $ACY$ respectively. Let us note that $SA = SP$ and $TA = TP$, from where the line $ST$ is the perpendicular bisector of $AP$. The problem then becomes equivalent to proving that $K$, $S$, and $T$ are collinear. But let us note that since $\angle BAX = \angle YAC$, by central angle we have that $\angle BSX = \angle YTC$, and since $BSX$ and $YTC$ are isosceles, these triangles are similar. Now, since $BX$ and $YC$ are parallel, we have by Thales that $\frac{KB}{KY} = \frac{KX}{KC}$. Then the dilation centered at $K$ that sends $B$ to $Y$ also sends $X$ to $C$. Let $T_1$ be the image of $S$ after applying this same dilation, we have that $BSX$ and $YT_1C$ are similar, and since both $T$ and $T_1$ are in the same half plane as $A$ with respect to $CY$ we have that $T = T_1$. Then $K$, $S$ and $T$ are collinear and $KA = KP$.