7. Given an equilateral $\triangle A B C$ with side length $\sqrt{3}$, point $D$ lies on the circumcircle of $\triangle A B C$, satisfying $A D=\sqrt{2}, P$ is a moving point on $\triangle A B C$. Then the range of $\overrightarrow{P A} \cdot \overrightarrow{P D}$ is $\qquad$
7. $\left[-\frac{2+\sqrt{3}}{8}, \frac{3+\sqrt{3}}{2}\right]$.

From the problem, we know that the circumradius of $\triangle ABC$ is 1, and let its center be $O$.
Given $AD=\sqrt{2}$, we know
$$
\begin{array}{l}
OA \perp OD, \text{ and }|\overrightarrow{OA}+\overrightarrow{OD}|=\sqrt{2}. \\
\text{Then } \overrightarrow{PA} \cdot \overrightarrow{PD} \\
=(\overrightarrow{OA}-\overrightarrow{OP}) \cdot(\overrightarrow{OD}-\overrightarrow{OP}) \\
=|\overrightarrow{OP}|^{2}-(\overrightarrow{OA}+\overrightarrow{OD}) \cdot \overrightarrow{OP} \\
=\left(\overrightarrow{OP}-\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OD})\right)^{2}-\frac{1}{4}(\overrightarrow{OA}+\overrightarrow{OD})^{2} \\
=|\overrightarrow{EP}|^{2}-\frac{1}{2},
\end{array}
$$

where $E$ is the midpoint of $AD$, i.e.,
$$
\overrightarrow{OE}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OD}).
$$

As shown in Figure 1, when point $P$ is on $AB$ and $EP \perp AB$, $|\overrightarrow{EP}|$ is minimized, at $\frac{\sqrt{3}-1}{4}$;

when point $P$ coincides with point $C$, $|\overrightarrow{EP}|$ is maximized, at $\sqrt{\frac{4+\sqrt{3}}{2}}$.

As point $P$ moves, $|\overrightarrow{EP}|$ changes continuously. Substituting into equation (1), we get
$$
\overrightarrow{PA} \cdot \overrightarrow{PD} \in\left[-\frac{2+\sqrt{3}}{8}, \frac{3+\sqrt{3}}{2}\right].
$$