7. In the tetrahedron $P-ABC$, $PB \perp AC$, $PH$ $\perp$ plane $ABC$ at point $H$, $H$ is inside $\triangle ABC$, $PB$ makes a $30^{\circ}$ angle with plane $ABC$, the area of $\triangle PAC$ is 1. When the dihedral angle $P-AC-B$ is $\qquad$, $S_{\triangle ABC}$ is maximized.
$7.60^{\circ}$.
From the problem, we know $\angle P B H=30^{\circ}$.
Extend $B H$, intersecting $A C$ at point $M$, and connect $P M$. Then $\angle P M B$ is the plane angle of the dihedral angle $P-A C-B$.
From $1=S_{\triangle P A C}=\frac{1}{2} A C \cdot P M \Rightarrow P M=\frac{2}{A C}$.
In $\triangle P M B$, by the Law of Sines, we have
$$
\begin{array}{l}
\frac{B M}{\sin \angle B P M}=\frac{P M}{\sin 30^{\circ}} \\
\Rightarrow B M=\frac{4 \sin \left(150^{\circ}-\angle P M B\right)}{A C} .
\end{array}
$$

Thus, $S_{\triangle A B C}=\frac{1}{2} A C \cdot B M$
$$
=2 \sin \left(150^{\circ}-\angle P M B\right) \text {. }
$$

Therefore, when $\angle P M B=60^{\circ}$, $S_{\triangle A B C}$ is maximized.