8. As shown in Figure 1, let $P$ be a point in the plane of $\triangle A B C$, and
$$
A P=\frac{1}{5} A B+\frac{2}{5} A C \text {. }
$$

Then the ratio of the area of $\triangle A B P$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{2}$
(C) $\frac{2}{5}$
(D) $\frac{2}{3}$
8. C.

As shown in Figure 4, extend $A P$ to $E$ such that $A P = \frac{1}{5} A E$.

Connect $B E$, and draw $E D$ $/ / B A$ intersecting the extension of $A C$ at point $D$.
From $A P = \frac{1}{5} A B + \frac{2}{5} A C^{\circ}$, we get $A C = C D$.
Therefore, quadrilateral $A B E D$ is a parallelogram.
Thus, $\frac{S_{\triangle A B P}}{S_{\triangle B E}} = \frac{1}{5}$.
$$
\text{Also, } \frac{S_{\triangle A B E}}{S_{\triangle A B C}} = \frac{\frac{1}{2} S_{\square A B S D}}{\frac{1}{4} S_{\square \triangle A B E D}} = 2 \text{, so } \frac{S_{\triangle A B P}}{S_{\triangle B C}} = \frac{2}{5} \text{. }
$$