4. Given that $\triangle A B C$ and $\triangle B C D$ are both equilateral triangles, and points $A$ and $D$ are on opposite sides of side $B C$. A line through $D$ intersects the extension of $A B$ at point $P$ and the extension of $A C$ at point $Q$. Connecting $P C$ and $Q B$ intersect at point $M$. Then $\angle B M C=(\quad)$.
(A) $105^{\circ}$
(B) $120^{\circ}$
(C) $145^{\circ}$
(D) $150^{\circ}$
4. B. As shown in Figure 2, it is easy to see that $\triangle P B D \backsim \triangle D C Q$.
$$
\text { Therefore, } \frac{P B}{D C}=\frac{B D}{C Q} \text {. }
$$

Since $D C=B D=B C$, we have
$$
\frac{P B}{B C}=\frac{B C}{C Q} \text {. }
$$

Given $\angle P B C=\angle B C Q=$
$120^{\circ}$, we can conclude
$$
\triangle P B C \backsim \triangle B C Q \Rightarrow \angle 2=\angle 1 \text {. }
$$

In $\triangle B P M$, we have
$$
\begin{array}{l}
\angle B M C=\angle 2+\angle P B M=\angle 1+\angle P B M \\
=\angle P B C=120^{\circ}
\end{array}
$$