Since $Q$ is on both circumcircles of $ABC$ and $AXY$, so there is a spiral similarity about $Q$ carrying one of these circles to another such that it carries $X$ to $B$, and $Y$ to $C$. Suppose this similarity carries $K$ to a point $T$. It is enough that we prove that the points $P, K, T$ are collinear, because then, considering the similarity of $QKT$ and $QXB$ we have:
$$\angle QXA = 180^\circ - \angle QXB = 180^\circ - \angle QKT = \angle QKP$$
Now we prove that $P, K, T$ are collinear:
The spiral similarity about $Q$, carries $X, Y, K$ to $B, C, T$ respectively. So the triangles $XYK$ and $BCT$ are similar. From the other hand: (let $m(AB)$ denote the measure of the arc $AB$)
$$\angle BXK = \angle BPK = \frac{m(YK)}{2} = \angle XYK \\ \angle XBK = \frac{m(KX)}{2} = \angle YXK$$

Thus the triangles $BXK$ and $XYK$ are similar, and hence $BXK$ and $BCT$ are similar. So $\frac{BX}{BK} = \frac{BC}{BT}$, also $\angle XBK = \angle CBT \Rightarrow \angle XBC = \angle KBT$. From which we deduce that $BKT$ and $BXC$ are similar. So $\angle BXC = \angle BKT$, and:
$$\Rightarrow \angle BKT + \angle BKP = \angle BXC + \angle BKP = 180^\circ \Rightarrow K,P,T \text{ are collinear}$$