Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\beta = \angle Q A C = \angle C B A$ and $\gamma = \angle P A B = \angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain
$$\frac{B P}{P M} = \frac{B P}{P A} = \frac{A Q}{Q C} = \frac{N Q}{Q C}.$$
Moreover,
$$\angle B P M = \beta + \gamma = \angle C Q N.$$
Hence the triangles $B P M$ and $N Q C$ are similar. This gives $\angle B M P = \angle N C Q$, so the triangles $B P M$ and $B S C$ are also similar. Thus we get
$$\angle C S B = \angle B P M = \beta + \gamma = 180^\circ - \angle B A C,$$
which completes the solution.

Figure 1

As in the previous solution, denote by $S$ the intersection point of the lines $B M$ and $N C$. Let moreover the circumcircle of the triangle $A B C$ intersect the lines $A P$ and $A Q$ again at $K$ and $L$, respectively (see Figure 2).

Note that $\angle L B C = \angle L A C = \angle C B A$ and similarly $\angle K C B = \angle K A B = \angle B C A$. It implies that the lines $B L$ and $C K$ meet at a point $X$, being symmetric to the point $A$ with respect to the line $B C$. Since $A P = P M$ and $A Q = Q N$, it follows that $X$ lies on the line $M N$. Therefore, using Pascal's theorem for the hexagon $A L B S C K$, we infer that $S$ lies on the circumcircle of the triangle $A B C$, which finishes the proof.

Figure 2