The given conditions show that plane $LMN$ is parallel to plane $ABC$ and the ratio of the distances from point $P$ to planes $LMN$ and $ABC$ is $1 : 2$. Let $H$ be the centroid of face $ABC$ in regular triangular pyramid $P$-$ABC$ and $PH$ intersects plane $LMN$ at point $K$. Then $PH \perp ABC$ and $PK \perp LMN$, and thus $PK = \frac{1}{2}PH$.

The regular triangular pyramid $P$-$ABC$ can be regarded as a regular tetrahedron. Let $O$ be the centre of its circumscribed sphere. Then $O$ is on $PH$ and by the properties of regular tetrahedron we know that $OH = \frac{1}{4}PH$. By combining $PK = \frac{1}{2}PH$ we know that $OK = OH$, that is, point $O$ is equally distant to planes $LMN$ and $ABC$. This shows that the section circle of the circumscribed sphere of this regular triangular pyramid intercepted by planes $LMN$ and $ABC$ is equal in size.

As a result, the area of the required cross section is equal to the area of the circumcircle of $\triangle ABC$, namely, $\pi \cdot \left(\frac{AB}{\sqrt{3}}\right)^2 = \frac{\pi}{3}$.