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
!
Figure 2