Example 3 In $\triangle A B C$, $\angle A B C=40^{\circ}$, $\angle A C B=30^{\circ}, P$ is a point on the bisector of $\angle A B C$, $\angle P C B=10^{\circ}$. Find the degree measure of $\angle P A B$.
Solution: As shown in Figure 3, take a point $D$ on the extension of $BA$ such that $BD = BC$. Connect $D \bar{r}^{\prime}$ and $NC$.

Since $BP$ bisects $\angle ABC$, point $D$ is symmetric to point $C$ with respect to $BP$. Therefore,
$$
PD = PC.
$$

Given $\angle DPC = 2(\angle PBC + \angle PCB) = 60^{\circ}$, we know that $\triangle PCD$ is an equilateral triangle. Therefore,
$$
PC = DC.
$$

In $\triangle ACD$, since $\angle ADC = 70^{\circ} = \angle DAC$, we have $AC = DC$. Therefore,
$$
AC = PC.
$$

In $\triangle PCA$, since $\angle PCA = 20^{\circ}$, we have $\angle PAC = 80^{\circ}$.
Thus, $\angle PAB = 30^{\circ}$.
Here, since $BP$ bisects $\angle ABC$, we consider taking a point $D$ on the extension of $BA$ such that $BD = BC$, making point $D$ the symmetric point of $C$ with respect to $BP$. This is the simplest and most basic method of taking a symmetric point.