4. In an equilateral $\triangle A B C$, $D$ and $E$ are points on $B C$ and $A C$ respectively, and $A E=C D$. Connecting $A D$ and $B E$ intersect at point $P$, and connecting $P C$. If $P B=2 P A$, then the degree of $\angle C P D$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
4. A.

It is easy to know that $\triangle A B E \cong \triangle C A D$.
Therefore, $\angle A B E=\angle C A D$.
$$
\begin{array}{l}
\text { Also, } \angle B P D=\angle A B E+\angle B A P \\
=\angle C A D+\angle B A P=\angle B A C=60^{\circ} .
\end{array}
$$

Draw $B Q \perp P D$ at point $Q$. Then
$$
\angle P B Q=30^{\circ}, P Q=\frac{1}{2} P B=A P \text {. }
$$

Thus, $A Q=A P+P Q=P B$.
Also, $\angle P B C=\angle B A Q$, so $\triangle A B Q \cong \triangle B C P$.
Therefore, $\angle B P C=\angle A Q B=90^{\circ}$.
Hence, $\angle C P D=\angle B P C-\angle B P D=30^{\circ}$.