5. As shown in Figure 1, let $P$ be a point inside $\triangle A B C$, and $A P$ $=\frac{2}{5} A B+\frac{1}{5} A C$. Then $\frac{S_{\triangle B P}}{S_{\triangle A B C}}=(\quad)$.
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{1}{4}$
(D) $\frac{1}{3}$
5.A.

As shown in Figure 6, let $A M=\frac{2}{5} A B$,
$$
A N=\frac{1}{5} A C \text {. }
$$

Then $A P=A M+A N$.
By the parallelogram rule, $N P / / A B$, so,
$$
\frac{S_{\text {SUHP }}}{S_{\triangle M K C}}=\frac{|A N|}{|A C|}=\frac{1}{5} .
$$