Example 5 In a regular pentagon $A B C D E$, there is a point $P$ inside. It is known that $\angle A B P=6^{\circ}, \angle A E P=12^{\circ}$. Find the degree measure of $\angle P A C$.
Explanation: As shown in Figure 4, construct the equilateral triangle $\triangle AEF$, and connect $BF$. It is easy to know that the interior angle of the regular pentagon $ABCDE$ is $108^{\circ}$. Therefore,
$$
\begin{array}{l}
\angle BAF = 108^{\circ} + 60^{\circ} \\
= 168^{\circ}.
\end{array}
$$

Since $AF = AE = AB$,
then
$$
\begin{array}{l}
\angle AFB = \angle ABF \\
= \frac{1}{2}(180^{\circ} - 168^{\circ}) \\
= 6^{\circ} = \angle ABP.
\end{array}
$$

Thus, points $B$, $P$, and $F$ are collinear.
$$
\begin{array}{l}
\text{Also, } \angle PEF = \angle AEP + \angle AEF \\
= 12^{\circ} + 60^{\circ} = 72^{\circ}, \\
\angle EFP = \angle AFE - \angle AFB \\
= 60^{\circ} - 6^{\circ} = 54^{\circ},
\end{array}
$$

then $\angle EPF = 180^{\circ} - 72^{\circ} - 54^{\circ} = \angle EFP$.
Therefore, $EP = EF = AE$.
At this point, one of the base angles of the isosceles $\triangle EAP$ is
$$
\angle EAP = \frac{1}{2}(180^{\circ} - 12^{\circ}) = 84^{\circ}.
$$

And $\angle EAC = 180^{\circ} - \angle AED$
$$
= 180^{\circ} - 108^{\circ} = 72^{\circ},
$$

thus $\angle PAC = 84^{\circ} - 72^{\circ} = 12^{\circ}$.