It suffices to show that $X$, $B$, $Y$, $Q$ are cyclic under the assumption that $Q$ is the intersection of the line $BD$ and $\Gamma_2$. There are six cases depending on the ordering of $P$, $D$, $B$, $Q$. Although we should consider all those six cases, we here give the proof only for the case where the order is $P$, $D$, $B$, $Q$.

Let $\angle ABP = \alpha$, $\angle DPX = \beta$. Since $AB // CD$, we have $\angle ABP = \angle BDC = \alpha$. For the circle $\Gamma_1$, the circumference angle for $PB$ equals $\frac{\pi}{2} - \alpha$; hence we have $(\frac{\pi}{2} - \alpha) + \angle PXB = \pi$. That is, $\angle PXB = \frac{\pi}{2} + \alpha$.

On the other hand, we have
$$\angle DYQ = \frac{1}{2} \angle DCQ = \frac{\pi}{2} - \alpha.$$
For the circle $\Gamma$, $\angle DPX = \angle DYX = \beta$. Now it suffices to show that $\angle DBX = \angle XYQ$.

For the triangle $DBX$, we have
$$\angle DBX = \pi - \beta - (\frac{\pi}{2} + \alpha) = \frac{\pi}{2} - \alpha - \beta.$$
Meanwhile, since $\angle DYQ = \frac{\pi}{2} - \alpha$ and $\angle DYX = \beta$, we have
$$$$
$\angle$ XYQ = $\angle$ DYQ - $\angle$ DYX = $\frac{\pi}{2} - \alpha - \beta. \quad \square$