Note that $(ABM)$ is tangent to $MP$. We fix triangle $AMN$ and move $B,C$ on $M,N$. $C\mapsto B\mapsto (AMB)\mapsto MP\mapsto P$ is projective. Consider the line $CP$, when $B=M$,$C=N$, and $P=N$, so by Steiner conic $CP$ passes through a fixed point. When $C=M$, $B=N$, so $CP$ become the tangent of $(AMN)$ at $M$. When $C=B=\infty_{MN}$, $CP$ becomes $A\infty_{MN}$. Therefore $CP$ passes through the fixed point $X$, which is the imtersection of $A\infty_{MN}$ and the tangent of $(AMN)$ at $M$. Similarly $BQ$ passes through a fixed point $Y$, and it is easy to see that $CX$ and $BY$ are symmetric wrt the perpendicular bisector of $MN$, so $\angle QBC=\angle PCB$ follows.

Let MP, NQ intersect (AMN) at Y and Z respectively; then by angle chasing, AZM ~ ANC and ANY ~ ABM, so if (AMN) intersects AB, AC at Y’ and Z’ then Y, Y’ are reflections over the perpendicular bisector of BC, as are Z, Z’. By Pascal on MAY’ZNM, the intersection of Y’Z and the tangent at M lies on line BQ, and similarly the intersection of YZ’ and the tangent at N lies on CP. Since these two intersection points are reflections over the perpendicular bisector of BC, by symmetry so are BQ and CP, which completes the proof.