4. As shown in Figure $4, \odot O$ is a unit circle.
$A B 、 C D$ are both its diameters, $\overparen{A D}=$ $60^{\circ}, P$ is a point on $\overparen{D B}$. Then $(P A+$ $P C)_{\text {max }}=$ $\qquad$ .$(P A+P C)_{\text {min }}$ $=$
$4.2 \sqrt{3}, 3$.
As shown in Figure 8, connect $B C$. It is easy to prove that
$\triangle A M O \subseteq \triangle C N B$.
Thus, $A M = C N, M O = N B$.
It is easy to prove that $\triangle A O M \backsim \triangle A P N$,
$\triangle C O N \backsim \triangle C P M$.
Therefore, $A M \cdot A P = A O \cdot A N$,
$C N \cdot C P = C O \cdot C M$.
$$
\begin{aligned}
\therefore & A M(P A + P C) \\
& = A O(A N + C M) \\
& = 1 \cdot (1 + O N + 1 + O M) = 1 + 1 + 1 = 3 .
\end{aligned}
$$

Thus, $P A + P C = \frac{3}{A M}$.
$$
\because \frac{\sqrt{3}}{2} \leqslant A M \leqslant 1, \therefore 3 \leqslant P A + P C \leqslant 2 \sqrt{3} \text{. }
$$