4. As shown in the figure, it is known that $A B C D E F$ is a regular hexagon, $M$ and $N$ are the midpoints of sides $C D$ and $D E$ respectively, and $A M$ intersects $B N$ at point $P$. Then $\angle A P B$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
4. (C).

As shown in the figure, connect $BD$ and $AC$, then $\angle BDN = \angle ACM = 90^{\circ}$, $BD = AC$, $DN = CM$, so $\triangle BDN \cong \triangle ACM$. Therefore, $\angle PND = \angle PMC$, which means points $P$, $N$, $D$, and $M$ are concyclic. Hence, $\angle APB = \angle MPN = 180^{\circ} - \angle MDN = 180^{\circ} - 120^{\circ} = 60^{\circ}$.