3. As shown in Figure 1, in $\triangle A B C$, $A B=A C, \angle A=40^{\circ}$, extend $A C$ to $D$ such that $C D=B C$, point $P$ is the incenter of $\triangle A B D$. Then $\angle B P C=(\quad)$.
(A) $145^{\circ}$
(B) $135^{\circ}$
(C) $120^{\circ}$
(D) $105^{\circ}$
3. (A).

Connect $P D$, by the property of the incenter we know $\angle B P D=90^{\circ}+\frac{1}{2} \angle A$.
Also, $\angle B C D=90^{\circ}+\frac{1}{2} \angle A$, thus
$\angle B P D=\angle B C D$.
Therefore, $B, D, C, P$ are concyclic.
It is easy to know $\angle D=35^{\circ}$, so, $\angle B P C=180^{\circ}-35^{\circ}=145^{\circ}$.