Example 8 As shown in Figure 8-5, in $\triangle A B C$, $B M: B C=1: 3, B N: B A=3: 5, A M$ intersects $C N$ at point $P$. If $\overrightarrow{B C}=\vec{a}, \overrightarrow{B A}=\vec{b}$, express $\overrightarrow{B P}$ in terms of $\vec{a}, \vec{b}$.
Solve: Since $B M: B C=1: 3, B N: B A=3: 5, \overrightarrow{B C}=\vec{a}, \overrightarrow{B A}=\vec{b}$, therefore $\overrightarrow{B M}=\frac{1}{3} \vec{a}, \overrightarrow{B N}=\frac{3}{5} \vec{b}$. Let $\overrightarrow{M P}=\alpha \overrightarrow{M A}$, then
$$
\begin{array}{l}
\overrightarrow{B P}=\overrightarrow{B M}+\overrightarrow{M P}=\frac{1}{3} \vec{a}+\alpha \overrightarrow{M A}=\frac{1}{3} \vec{a}+\alpha\left(\vec{b}-\frac{1}{3} \vec{a}\right)= \\
\frac{1-\alpha}{3} \vec{a}+\alpha \vec{b} .
\end{array}
$$

Let $\overrightarrow{N P}=\beta \overrightarrow{N C}$, then $\overrightarrow{B P}=\overrightarrow{B N}+\overrightarrow{N P}=\frac{3}{5} \vec{b}+\beta(\overrightarrow{B C}-$ $\overrightarrow{B N})=\frac{3-3 \beta}{5}+\beta \vec{a}$,
By the fundamental theorem of plane vectors, we know
$$
\left\{\begin{array}{l}
\frac{1-\alpha}{3}=\beta \\
\alpha=\frac{3-3 \beta}{5}
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\alpha=\frac{1}{2}, \\
\beta=\frac{1}{6} .
\end{array}\right.\right.
$$

Therefore, $\overrightarrow{B P}=\frac{1}{6} \vec{a}+\frac{1}{2} \vec{b}$.