4. As shown in Figure 2, given that $P$ is a point inside a square $A B C D$ with side length $a$, and $\triangle P B C$ is an equilateral triangle. Then the circumradius of $\triangle P A D$ is ( ).
(A) $a$
(B) $\sqrt{2} a$
(C) $\frac{\sqrt{3}}{2} a$
(D) $\frac{1}{2} a$
4.A.

As shown in Figure 6, let the circumcenter of $\triangle P A D$ be $O$, and connect $O A$, $O P$, and $O D$.

Since $\triangle P A D$ and $\triangle O A D$ are both isosceles triangles, $O P$ is the perpendicular bisector of $A D$. Therefore, $O P \parallel A B$.
Given $\angle A B P=30^{\circ}$, then $\angle O P B=150^{\circ}$.
However, $\angle A P B=\angle P A B=75^{\circ}$, thus,
$$
\angle O P A=\angle O A P=75^{\circ} \text {. }
$$

Therefore, in $\triangle O A P$ and $\triangle B A P$, the two angles are equal, and they share the common side $A P$. Hence,
$$
\triangle O A P \cong \triangle B A P \Rightarrow O A=A B=a .
$$