Example 9 In $\triangle A B C$, $\angle A B C=\angle A C B=40^{\circ}$, $P$ is a point inside the triangle; $\angle P A C=20^{\circ}$, $\angle P C B=30^{\circ}$. Find the degree measure of $\angle P B C$.
Solution: As shown in Figure 9, let point $D$ be the symmetric point of point $C$ with respect to $AP$. Connect $DA$, $DB$, $DC$, and $DP$.
Given $\angle PAC=20^{\circ}$, $\angle PCA=10^{\circ}$, we have
$\angle DAC=40^{\circ}$, $\angle PDA=\angle PCA=10^{\circ}$,
thus $\triangle PDC$ is an equilateral triangle.
Given $\angle ABC=\angle ACB=40^{\circ}$, we have
$AC=AB=AD$.
Given $\angle BAD=60^{\circ}$, we know $\triangle ABD$ is an equilateral triangle. Therefore, $\angle DBC=60^{\circ}-\angle ABC=20^{\circ}$.

Given $\angle PCB=30^{\circ}$, point $P$ is symmetric to point $D$ with respect to $BC$.
Hence, $\angle PEC=\angle DBC=20^{\circ}$.