The answer is $\boxed{\angle QMP=90-\frac{\angle BAC}{2}}$ First by the problem condition one can see that $QF,PE,BC$ are all concurrent at $D$ and $AQBD,APCD$ are cyclic, now let $EF \cap BC=G$ and let $EF$ hit $(ABC)$ at $X,Y$ where the points are in this order $G.X,F,E,Y$, now its clear that $-1=(G, D; B, C)$ so by McLaurin $GX \cdot GY=GB \cdot GC=GD \cdot GM$ so $DXYM$ is cyclic and now by PoP $QF \cdot FD=AF \cdot FB=XF \cdot FY$ and $PF \cdot FB=AF \cdot FC=XF \cdot FY$ which gives that $QXDMYP$ is cyclic so $\angle QMP=\angle FDE=90-\frac{\angle BAC}{2}$ thus we are done