3. As shown in Figure $1, \odot O$'s chord $A B$ intersects with $C D$ at point $P, A O$ intersects with $C D$ at point $Q$. If $A O \perp C D$, $\angle A O D=2 \angle P A C$, then among the following four conclusions, the correct one is ( ).
(A) $D, O, B$ are collinear
(B) $A, D, O, P$ are concyclic
(C) $P$ is the incenter of $\triangle A O C$
(D) $Q B \perp O C$
3.B.

Let $\angle A O D=\alpha$. By the perpendicular diameter theorem, we have $\angle A O C=\alpha$.
Thus, $\angle A B C=\frac{1}{2} \angle A O C=\frac{\alpha}{2}=\angle P A C$.
Therefore, $C$ is the midpoint of $\overparen{A B}$, meaning points $A, C$ trisect $\overparen{D B}$. Hence,
$$
\begin{array}{l}
\angle A P D=\angle P A C+\angle P C A \\
=\frac{1}{2} \angle B O C+\frac{1}{2} \angle A O D=\frac{\alpha}{2}+\frac{\alpha}{2} \\
=\alpha=\angle A O D .
\end{array}
$$

Therefore, points $A, D, O, P$ are concyclic.