3. In $\triangle A B C$, it is known that $A B=A C, D$ is the midpoint of side $B C$, $B E \perp A C$ at point $E, B E$ intersects $A D$ at point $P$. If $B P=3, P E=1$, then $A E=(\quad)$.
(A) $\frac{\sqrt{6}}{2}$
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) $\sqrt{6}$
3. B.

Since $A D \perp B C, B E \perp A C$, therefore, $P, D, C, E$ are concyclic.
Thus, $B D \cdot B C = B P \cdot B E = 12$.
Also, $B C = 2 B D$, so $B D = \sqrt{6}$.
Hence, $D P = \sqrt{3}$.
By $\triangle A E P \sim \triangle B D P \Rightarrow \frac{A E}{B D} = \frac{P E}{D P}$
$\Rightarrow A E = \frac{P E}{D P} \cdot B D = \frac{1}{\sqrt{3}} \times \sqrt{6} = \sqrt{2}$.