2. Let $D$ be a point on side $A B$ of $\triangle A B C$, and $P$ be a point inside $\triangle A B C$ such that
$$
\begin{array}{l}
\boldsymbol{A D}=\frac{3}{4} \boldsymbol{A B}, \boldsymbol{A P}=\boldsymbol{A} D+\frac{2}{5} \boldsymbol{B} C . \\
\text { Then } \frac{S_{\triangle P D}}{S_{\triangle M B C}}=(\quad) .
\end{array}
$$
(A) $\frac{3}{10}$
(B) $\frac{2}{5}$
(C) $\frac{7}{15}$
(D) $\frac{8}{15}$
2. A.

Connect $P D$, then $D P=\frac{2}{5} B C$.
Therefore, $D P / / B C, \angle A D P=\angle B$.
$$
\text { Hence } \frac{S_{\triangle A P D}}{S_{\triangle A B C}}=\frac{\frac{1}{2}|A D| \cdot|D P| \sin \angle A D P}{\frac{1}{2}|A B| \cdot|B C| \sin B}=\frac{3}{10} \text {. }
$$