4. A. As shown in Figure $1, \odot O$'s diameter $A B$ intersects with chord $C D$ at point $P$, forming an angle of $45^{\circ}$. If $P C^{2}+P D^{2}=8$, then the radius of $\odot O$ is ( ).
(A) $\sqrt{2}$
(B) 2
(C) $2 \sqrt{2}$
(D) 4
4. A. B.

Draw $O H \perp C D$, with the foot of the perpendicular at $H$. Without loss of generality, assume point $H$ lies on segment $C P$. Then $\angle O P H=45^{\circ}$.
Thus, $O H=H P=\frac{1}{2}(P C-P D)$.
By $C H=\frac{1}{2}(P C+P D)$, in the right triangle $\triangle O H C$, applying the Pythagorean theorem, we get
$$
\begin{array}{l}
O C^{2}=O H^{2}+C H^{2}=\frac{1}{2}\left(P C^{2}+P D^{2}\right)=4 \\
\Rightarrow O C=2
\end{array}
$$