Example 5 Let $P$ be a point inside $\triangle A B C$ such that $\angle P B C = 30^{\circ}, \angle P B A = 8^{\circ}$, and $\angle P A B = \angle P A C = 22^{\circ}$. Question: What is the measure of $\angle A P C$ in degrees?
Solve as shown in Figure 7, take
the circumcenter $D$ of $\triangle P B C$, and
connect $D P, D C, D B$.
Then $D P=D C=D B$,
$\angle P D C$
$=2 \angle P B C=60^{\circ}$.
Therefore, $\triangle D P C$ is an equilateral triangle.
Thus, $P C=D P=D B, \angle D P C=60^{\circ}$.
It is easy to see that $A B>A C$.
Therefore, we can take a point $E$ on $A B$ such that $A E=A C$, and connect $E P$.
It is easy to prove that $\triangle A P E \cong \triangle A P C$.
Thus, $P E=P C, \angle A P E=\angle A P C$.
Hence $P E=D P=D B$,
$\angle E P B=\angle A P B-\angle A P E$
$=150^{\circ}-\angle A P C$.
Also, $\angle D B P=\angle D P B$
$=360^{\circ}-\angle A P B-\angle A P C-\angle D P C$
$=150^{\circ}-\angle A P C$,
so $\angle E P B=\angle D B P$.
Therefore, $P E \parallel D B$.
Thus, quadrilateral $B D P E$ is a rhombus, and $E B=E P$.
Therefore, $\angle E P B=\angle E B P=8^{\circ}$.
Hence $\angle A P C=\angle A P E$
$=\angle A P B-\angle E P B=142^{\circ}$.