4. As shown in Figure 2, $A B C D$ is a square with an area of 1, and $\triangle P B C$ is an equilateral triangle. Then the area of $\triangle B P D$ is ( ).
(A) $\frac{\sqrt{3}-1}{2}$
(B) $\frac{2 \sqrt{3}-1}{8}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{3}-1}{4}$
4.D.

As shown in Figure 6, construct $P E \perp C D$ at $E$, and $P F \perp B C$ at $F$. Then $P E=\frac{1}{2} C P$ $=\frac{1}{2}, P F=P C \sin 60^{\circ}=\frac{\sqrt{3}}{2}$. Therefore,
$$
\begin{array}{l}
S_{\triangle P P D}=S_{\text {quadrilateral } B C D P}-S_{\triangle B C D} \\
=S_{\triangle C P}+S_{\triangle A C D}-S_{\triangle A C D} \\
=\frac{1}{2} B C \cdot P F+\frac{1}{2} C D \cdot P E-\frac{1}{2} B C \cdot C D \\
=\frac{\sqrt{3}}{4}+\frac{1}{4}-\frac{1}{2}=\frac{\sqrt{3}-1}{4} \text {. } \\
\end{array}
$$