Denote $\angle MAC = \alpha$ and $\angle ACB = \beta$. Let $J$ be the incenter of $\triangle AMC$. It follows that $\angle AJM = 90^\circ + \beta/2$ and $\angle ABM = 90^\circ - \beta/2$, which implies that the quadrilateral $ABMJ$ is inscribed in a circle $k_1$. Analogously we have that the quadrilateral $MNCJ$ is inscribed in a circle $k_2$ which in fact is the the circumcircle of $\triangle MCN$. Therefore $P$ is a point of $k_2$. Since $QA \cdot QB = QC \cdot QP$, we conclude that $Q$ has one and the same degree with respect to $k_1$ and $k_2$. Hence $Q$ lies on $JM$, which is the angular bisector of $\angle BNC$. This implies the desired equality $\angle QMB = \angle QMN$.