Bulgaria

Denote $\angle MAC=\alpha$ and $\angle ACB=\beta$. Let $J$ be the incenter of $△AMC$. 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 $△MCN$. 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 $\angle QMB=\angle QMN$.