4. As shown in Figure 2, equilateral $\triangle A B C$ is inscribed in $\odot O, P$ is any point on the minor arc $\overparen{B C}$, connect $P A, P B, P C, P A$ intersects $B C$ at point $D$, let the area of quadrilateral $A B P C$ be $4 \sqrt{3}$. Then ( ).
(A) $P D \leqslant 1$
(B) $P D \geqslant 1$
(C) $P D \leqslant \sqrt{3}$
(D) $P D \geqslant \sqrt{3}$
4. A.

As shown in Figure 5, $\triangle A B P$ is rotated counterclockwise by $60^{\circ}$ around point $A$ to $\triangle A C E$.

It is easy to see that points $P$, $C$, and $E$ are collinear, and $\triangle A P E$ is an equilateral triangle.
$$
\begin{array}{l} 
\text { Also, } S_{\triangle A P E} \\
= S_{\text {quadrilateral } A B P C}=4 \sqrt{3} \\
\Rightarrow \frac{\sqrt{3}}{4} P A^{2}=4 \sqrt{3} \Rightarrow P A=4 .
\end{array}
$$

Thus, $P B+P C=C E+P C=P E=P A=4$.
Since $\triangle B P D \backsim \triangle A P C \Rightarrow \frac{B P}{P D}=\frac{A P}{P C}$
$$
\Rightarrow P D=\frac{P B \cdot P C}{P A} \leqslant \frac{(P B+P C)^{2}}{4 P A}=\frac{P A}{4}=1 \text {. }
$$