4. Given that the radius of $\odot O$ is $R$, $C, D$ are two points on the circumference of the circle on the same side of the diameter $A B$, the degree measure of $\overparen{A C}$ is $96^{\circ}$. The degree measure of $\overparen{B D}$ is $36^{\circ}$, and a moving point $P$ is on $A B$. Then the minimum value of $C P+P D$ is $\qquad$.
4. $\sqrt{3} R$.

If $D^{\prime}$ is the point symmetric to $D$ with respect to line $A B$, and $C D^{\prime}$ intersects $A B$ at point $P$, then point $P$ minimizes $(P+P D)$ (proof omitted).

Given $36^{\circ}$, thus the measure of $\overparen{C D}$ is $180^{\circ}-96^{\circ}-36^{\circ}=48^{\circ}$. Therefore, the measure of $\overparen{C D^{\prime}}$ is $48^{\circ}+2 \times 36^{\circ}=120^{\circ}$. Hence, $\angle C O D^{\prime}=120^{\circ}$, and it is easy to see that
$$
C P+P D=C P+P D^{\prime}=C D^{\prime}=\sqrt{3} R .
$$