8. Let the base edge length of the regular tetrahedron $P-ABC$ be $a$. A plane through $A$ parallel to $BC$ intersects the side face $PBC$ perpendicularly, and the angle between the intersecting plane and the base is $30^{\circ}$. Find the area of the intersecting plane.
8. As shown in the figure, let the section intersect $PB$ at $E$, $PC$ at $F$, the midpoint of $EF$ be $H$, and the midpoint of $BC$ be $D$.

Since $BC$ // section $AEF$, it follows that $BC \parallel EF$, thus $H$ lies on $PD$,
and since $EF \perp PD$ and $EF \perp AD$, we know that $EF \perp$ plane $PAD$.
Draw a line $l \parallel EF$ through $A$, it can be proven that $l$ is the intersection line of section $AEF$ and plane $ABC$,
and $l \perp$ plane $PAD$, hence $l \perp AH$ and $l \perp AD$.
Therefore, the angle between section $AEF$ and the base $ABC$ is $\angle HAD = 30^\circ$.
Since section $AEF \perp$ plane $PBC$ and plane $PBC \perp$ plane $PAD$,
and section $AEF \cap$ plane $PAD = AH$, it follows that $AH \perp$ plane $PBC$.
Thus, $AH \perp PD$ and $AH \perp EF$.
In the right triangle $\triangle AHD$, $AH = AD \cdot \cos 30^\circ = \frac{\sqrt{3}}{2} AD = \frac{3}{4} a$,
$$
\begin{array}{c}
HD = AD \cdot \sin 30^\circ = \frac{1}{2} AD = \frac{\sqrt{3}}{4} a, \\
PD = \frac{1}{3} AD \div \cos 60^\circ = \frac{\sqrt{3}}{3} a, \text{ so } PH = PD - HD = \frac{\sqrt{3}}{3} a - \frac{\sqrt{3}}{4} a = \frac{\sqrt{3}}{12} a,
\end{array}
$$

Therefore, $\frac{EF}{BC} = \frac{PH}{PD} = \frac{1}{4}$, so $EF = \frac{a}{4}$. Thus, $S_{\text{section } AEF} = \frac{1}{2} \cdot AH \cdot EF = \frac{3}{32} a^2$.