7. In the quadrilateral pyramid $P-ABCD$, plane $PAD \perp$ plane $ABCD$, $\triangle PAD$ is an equilateral triangle with side length 2, and the base $ABCD$ is a rhombus with $\angle BAD=60^{\circ}$. Then the size of the acute dihedral angle formed by plane $PAD$ and plane $PBC$ is $\qquad$
$7.45^{\circ}$.
Take the midpoint $E$ of $A D$, and connect $P E$, $B E$, $B D$.
Since quadrilateral $A B C D$ is a rhombus, $\angle B A D=60^{\circ}$
$\Rightarrow \triangle A B D$ is an equilateral triangle
$\Rightarrow A D \perp B E$.
Similarly, $A D \perp P E$.
Since plane $P A D \perp$ plane $A B C D$, and $P E$ is in plane $P A D$, thus, $P E \perp$ plane $A B C D$.
Since $E B$ is in plane $A B C D$, thus, $P E \perp B E$.
Hence, $E A$, $E B$, and $E P$ are mutually perpendicular.
Establish a spatial rectangular coordinate system $E-x y z$ with $E$ as the origin as shown in Figure 3.
By the problem statement,
$$
\begin{array}{l}
C(-2, \sqrt{3}, 0), P(0,0, \sqrt{3}) \text {. } \\
\overrightarrow{P B}=(0, \sqrt{3},-\sqrt{3}) \text {, } \\
\overrightarrow{P C}=(-2, \sqrt{3},-\sqrt{3}) \text {. } \\
\end{array}
$$

Then $E(0,0,0), B(0, \sqrt{3}, 0)$,

Thus, $\overrightarrow{E B}=(0, \sqrt{3}, 0)$,

Let $\boldsymbol{n}=(x, y, z)$ be a normal vector of plane $P B C$.
$$
\text { By }\left\{\begin{array}{l}
\boldsymbol{n} \cdot \overrightarrow{P B}=\sqrt{3} y-\sqrt{3} z=0, \\
\boldsymbol{n} \cdot \overrightarrow{P C}=-2 x+\sqrt{3} y-\sqrt{3} z=0,
\end{array}\right.
$$

Take $y=1$, then $\boldsymbol{n}=(0,1,1)$.
Since $\overrightarrow{E B}$ is a normal vector of plane $P A D$,
$$
\begin{array}{l}
\Rightarrow \cos \langle\overrightarrow{E B}, \boldsymbol{n}\rangle=\frac{\overrightarrow{E B} \cdot \boldsymbol{n}}{|\overrightarrow{E B}||\boldsymbol{n}|}=\frac{\sqrt{3}}{\sqrt{3} \sqrt{2}}=\frac{\sqrt{2}}{2} \\
\Rightarrow\langle\overrightarrow{E B} \cdot \boldsymbol{n}\rangle=45^{\circ} .
\end{array}
$$

Therefore, the size of the acute dihedral angle formed by plane $P A D$ and plane $P B C$ is $45^{\circ}$.