3. As shown in Figure 1, line segment
$A B$ intersects $C D$ at point $O, P$
is a point inside $\angle A O D$,
$\angle A C P=2 \angle P C D$,
$\angle A B P=2 \angle P B D$,
$\angle A=39^{\circ}, \angle D=27^{\circ}$.
Then $\angle P$ is ( ).
(A) $30^{\circ}$
(B) $31^{\circ}$
(C) $32^{\circ}$
(D) $33^{\circ}$
3. B.

Let $\angle P C D=\alpha, \angle P B D=\beta$.
In $\triangle O A C, \triangle O B D$, by the exterior angle theorem of a triangle, we have
$$
\begin{array}{l}
\angle A+\angle A C O=\angle A O D=\angle D+\angle O B D \\
\Rightarrow 39^{\circ}+3 \alpha=27^{\circ}+3 \beta .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\angle A+\angle A C P=\angle P+\angle A B P \\
\Rightarrow 39^{\circ}+2 \alpha=\angle P+2 \beta .
\end{array}
$$

Solving equations (1) and (2) simultaneously, we get $\angle P=31^{\circ}$.