3. As shown in Figure $2, \triangle A B C$'s exterior angle $\angle A C D$'s angle bisector $C P$ intersects with the angle bisector $B P$ of the interior angle $\angle A B C$ at point $P$. If $\angle B P C=\alpha$, then $\angle C A P=$ $\qquad$ (express in terms of $\alpha$).
Figure 2
$$
\text { 3. } 90^{\circ}-\alpha \text {. }
$$

It is easy to know that $P$ is the excenter of $\triangle A B C$. Then
$$
\begin{array}{l}
\angle B P C=\angle P C D-\angle P B C=\frac{1}{2} \angle B A \\
\Rightarrow \angle B A C=2 \alpha \\
\Rightarrow \angle C A P=\frac{180^{\circ}-2 \alpha}{2}=90^{\circ}-\alpha .
\end{array}
$$