3. As shown in Figure $2, \odot O$'s diameter $A B$ intersects with the line of chord $C D$ (not a diameter) at a point $P$ outside the circle, $P A=A B=4$. Then the range of $P D$ is ( ).
(A) $4<P D<8$
(B) $4 \sqrt{2}<P D<8$
(C) $4<P D<4 \sqrt{2}$
(D) $4 \sqrt{2} \leqslant P D \leqslant 8$
3. B.

As shown in Figure 5, connect $O D$. In $\triangle P O D$, we have $P O - O D \angle P E C = \angle E D C$.
Therefore, $P D > P E$, which means $P D > 4 \sqrt{2}$.
In conclusion, $4 \sqrt{2} < P D < 8$.