Example 2 In $\triangle A B C$, $\angle A B C=\angle A C B=40^{\circ}$, $P$ is a point inside the triangle, $\angle P C A=\angle P A C=20^{\circ}$. Find the degree measure of $\angle P B C$.
Solution: As shown in Figure 2, construct a regular $\triangle DAC$ outside $\triangle ABC$ with $AC$ as one side. Connect $DP$.

Given $\angle PCA = \angle PAC = 20^{\circ}$, we know $PA = PC$. Point $A$ is symmetric to point $C$ with respect to $PD$. Therefore,
$$
\angle PDA = \frac{1}{2} \angle ADC = 30^{\circ}.
$$

Given $\angle ACB = \angle ABC = 40^{\circ}$, we know
$$
AB = AC = AD.
$$

It is easy to see that $\angle PAD = 80^{\circ} = \angle PAB$, so point $B$ is symmetric to point $D$ with respect to $PA$. Thus, $\angle PBA = \angle PDA = 30^{\circ}$.
Therefore, $\angle PBC = 10^{\circ}$.