In the proof, we will use $\angle(a, b)$ to denote the directed angle between line $a$ and line $b$. Let $D$ be the point on segment $AB$ satisfying $BD = \lambda \cdot BA$. We will show: either $Q = D$, or else $\angle(DQ, QB) = \angle(AB, BC)$; both cases guarantee that the point $Q$ moves on a certain circle passing through the point $D$ and tangent to line $BC$ at the point $B$. This is exactly the statement we wish to prove.

Denote the circumcircles of triangle $AMP$ and triangle $BMC$ by $\omega_A, \omega_B$ respectively. The three lines $AP$, $BC$, $MQ$ are pairwise the common radical axes of the three circles $\Gamma, \omega_A, \omega_B$, so these three lines are pairwise parallel, or else they are concurrent at a point $X$.

First suppose these three lines are mutually parallel, as in Figure 1. Then the three segments $AP$, $QM$, $BC$ have a common perpendicular bisector; reflecting through this perpendicular bisector maps segment $CP$ to segment $BA$, and sends $M$ to $Q$. Therefore $Q$ lies on segment $AB$, and $BQ/AB = CM/CP = BD/AB$; hence $Q = D$.

Next suppose the three lines $AP$, $QM$, $BC$ are concurrent at $X$, as in Figure 2. Applying Miquel's theorem in triangle $XPC$, we obtain that the four points $A, B, Q, X$ all lie on the same circle $\Gamma$. Let $Y$ be the reflection of $X$ over the perpendicular bisector of $AB$. It is easy to see that $Y$ also lies on $\Gamma$, and that $\triangle YAB$ and $\triangle XBA$ are congruent. Since $\triangle XPC$ is similar to $\triangle XBA$, it is also similar to $\triangle YAB$.

Since $BD/BA = CM/CP = \lambda$, we know that $D$ and $M$ correspond to each other in the similar triangles $YAB$, $XPC$ respectively. Moreover, since $\triangle YAB$ and $\triangle XPC$ have the same orientation, we have $\angle (MX, XP) = \angle (DY, YA)$. On the other hand, since $A, Q, X, Y$ all lie on $\Gamma$, we obtain $\angle (QY, YA) = \angle (MX, XP)$. Therefore $\angle (QY, YA) = \angle (DY, YA)$, that is, the three points $Y, D, Q$ are collinear.

Finally, we have $\angle (DQ, QB) = \angle (YQ, QB) = \angle (YA, AB) = \angle (AB, BX) = \angle (AB, BC)$, which is exactly what we wanted. Q.E.D.