Example 8 In $\triangle A B C$, $\angle A B C=60^{\circ}$, $\angle A C B=40^{\circ}, P$ is a point inside the triangle, $\angle P B C=20^{\circ}$, $\angle P C B=10^{\circ}$. Find the degree measure of $\angle P A B$.
Solution: As shown in Figure 8, let point $D$ be the reflection of point $B$ over $PC$. Connect $DA$, $DB$, $DC$, and $DP$.
In $\triangle BCD$, given $\angle DCB = 20^{\circ}$, we have
$\angle BDC = 80^{\circ} = \angle BAC$.
Thus, points $A$, $D$, $B$, and $C$ are concyclic.
Since $DC$ is parallel to $\angle ACB$, we have $DA = DB$.
It is easy to see that $\triangle PBD$ is an equilateral triangle, so
$DP = DB$.
Therefore, $DP = DA = DB$,

which means point $D$ is the circumcenter of $\triangle PAB$.
Hence, $\angle PAB = \frac{1}{2} \angle PDB = 30^{\circ}$.
Here, the reflection of point $B$ over $PC$, point $D$, happens to lie on the circumcircle of $\triangle ABC$, allowing the properties of a cyclic quadrilateral to play a role in the solution. It is evident that when choosing a reflection point, it is ideal if it can be placed on the circumcircle of a certain triangle.