7. Given that $D$ is a point on side $A B$ of $\triangle A B C$, and $P$ is a point inside $\triangle A B C$, and it satisfies
$$
\begin{array}{l}
\overrightarrow{A D}=\frac{3}{4} \overrightarrow{A B}, \overrightarrow{A P}=\overrightarrow{A D}+\frac{2}{5} \overrightarrow{B C} . \\
\text { Then } \frac{S_{\triangle A P D}}{S_{\triangle A B C}}=
\end{array}
$$
$7 \cdot \frac{3}{10}$.
Notice,
$$
\begin{array}{l}
\overrightarrow{D P}=\frac{2}{5} \overrightarrow{B C} \Rightarrow D P / / B C \\
\Rightarrow \angle A D P=\angle B \\
\Rightarrow \frac{S_{\triangle M P D}}{S_{\triangle A B C}}=\frac{\frac{1}{2}|A D||D P| \sin \angle A D P}{\frac{1}{2}|A B||B C| \sin B} \\
\quad=\frac{3}{4} \times \frac{2}{5}=\frac{3}{10} .
\end{array}
$$