2. As shown in Figure 1, $P$ is a point inside $\triangle A B C$, and satisfies $\boldsymbol{A P}=$ $\frac{2}{5} A B+\frac{1}{5} A C$. Then the ratio of the area of $\triangle P B C$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{5}$
(D) $\frac{2}{5}$
2.D.

As shown in Figure 6, let $A M=\frac{2}{5} A B$ and $A N=\frac{1}{5} A C$, then $A P=A M + A N$. By the parallelogram rule, we have
$$
M P \parallel A C, N P \parallel A B \text{. }
$$

Draw a line through point $P$ parallel to $B C$, intersecting $A B$ and $A C$ at points $B^{\prime}$ and $C^{\prime}$, respectively. Then the ratio of the heights from point $P$ to side $B C$ in $\triangle P B C$ and from point $A$ to side $B C$ in $\triangle A B C$ is $\frac{B^{\prime} B}{A B}$ (or $\frac{C^{\prime} C}{A C}$), denoted as $\lambda(0<\lambda<1)$. Therefore, $\frac{S_{\triangle P B C}}{S_{\triangle A B C}}=\lambda$.
By the properties of similar triangles, we have $\frac{M B^{\prime}}{N P}=\frac{M P}{N C^{\prime}}$, which means
$$
\frac{A B-A M-B^{\prime} B}{A M}=\frac{A N}{A C-A N-C^{\prime} C} .
$$

Thus, $\frac{\frac{3}{5} A B-B^{\prime} B}{\frac{2}{5} A B}=\frac{\frac{1}{5} A C}{\frac{4}{5} A C-C^{\prime} C}$.
Therefore, $\frac{3}{2}-\frac{5}{2} \lambda=\frac{\frac{1}{5}}{\frac{4}{5}-\lambda}$.
Solving for $\lambda$, we get $\lambda=1$ (discard), $\lambda=\frac{2}{5}$.