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