Example 4 As shown in Figure 5, the area of triangle $ABC$ is $1, BD: CD=2: 1, E$ is the midpoint of $AC$, and $AD$ intersects $BE$ at point $P$. Find the area of quadrilateral $PDCE$.
Let $S_{\triangle C E P}=x, S_{\triangle C D P}=y$.
From the problem, we have $\left\{\begin{array}{l}2 x+y=\frac{1}{3} \\ x+3 y=\frac{1}{2}\end{array}\right.$, solving this gives $\left\{\begin{array}{l}x=\frac{1}{10} \\ y=\frac{2}{15}\end{array}\right.$. Therefore, the area of quadrilateral $P D C E$ is $\frac{7}{30}$.