3. Given $P$ is a point inside isosceles $\triangle A B C$, $A B=$ $B C, \angle B P C=108^{\circ}, D$ is the midpoint of side $A C$, and $B D$ intersects $P C$ at point $E$. If $P$ is the incenter of $\triangle A B E$, then $\angle P A C=$ $\qquad$
$3.48^{\circ}$.
As shown in Figure 4, from the problem statement we have
$$
\begin{array}{l}
\angle P E A=\angle P E B \\
=\angle C E D=\angle A E D .
\end{array}
$$
$$
\text { and } \angle P E A+\angle P E B+
$$
$\angle A E D=180^{\circ}$, thus,
$$
\begin{array}{l}
\angle P E A=\angle P E B \\
=\angle C E D=\angle A E D \\
=60^{\circ} .
\end{array}
$$

Therefore, $\angle P C A=30^{\circ}$.
Since $\angle B P C=108^{\circ}$, we have,
$$
\begin{array}{l}
\angle P B E=12^{\circ} \Rightarrow \angle A B D=24^{\circ} \\
\Rightarrow \angle B A D=90^{\circ}-24^{\circ}=66^{\circ} \\
\Rightarrow \angle P A E=\frac{1}{2}(\angle B A D-\angle C A E)=18^{\circ} \\
\Rightarrow \angle P A C=\angle P A E+\angle C A E=48^{\circ} .
\end{array}
$$