IRL ABooklet

Let $T$ be the intersection point of the tangents to $\Omega$ at $A$ and $B$. Because of the right angles at $A$ and $B$, the points $A$, $M$, $B$, $T$ all lie on the circle with diameter $TM$. Because $|TA| = |TB|$ and these lines are tangents to $\Omega_A$ and $\Omega_B$, respectively, $T$ must be on the radical axis of $\Omega_A$ and $\Omega_B$, i.e. on the line $PQ$. Hence, $\angle PQM$ is a right angle iff $Q$ is on the circle with diameter $TM$.

To show this, first note that $\angle AMT = \frac{1}{2}\angle AMB = \angle TAB$ by the Alternate Segment Theorem in circle $\Omega$ with centre $M$ and that $\angle TAB = \angle AQP$ by the Alternate Segment Theorem in circle $\Omega_A$. Hence, $\angle AMT = \angle AQT$ which implies that $A$, $Q$, $M$, $T$ are concyclic. As $MT$ is a diameter of this circle and $P$ is on $QT$, it follows that $\angle PQM$ is a right angle.