4. Given a regular tetrahedron $P-ABC$ with the center of the base being $O$, a moving plane through $O$ intersects the lateral edges or their extensions of the tetrahedron at points $S, R, Q$. If
$$
\begin{array}{l}
\overrightarrow{P S}=m \overrightarrow{P A}, \overrightarrow{P R}=n \overrightarrow{P B}, \overrightarrow{P Q}=k \overrightarrow{P C}, \\
\text { then } \frac{m n k}{m n+n k+k m}=
\end{array}
$$
4. $\frac{1}{3}$.

Since points $S, R, Q, O$ are coplanar, we have
$$
\overrightarrow{P O}=x \overrightarrow{P S}+y \overrightarrow{P R}+z \overrightarrow{P Q},
$$

and $x+y+z=1$.
Thus, $\overrightarrow{P O}=x m \overrightarrow{P A}+y n \overrightarrow{P B}+z k \overrightarrow{P C}$.
From the fact that the center of the base of the regular tetrahedron $P-A B C$ is $O$, we get
$$
\overrightarrow{P O}=\frac{1}{3}(\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}).
$$

Since $\overrightarrow{P A}, \overrightarrow{P B}, \overrightarrow{P C}$ are not coplanar, we have
$$
m x=n y=k z=\frac{1}{3}.
$$
$$
\begin{array}{l}
\text { Therefore, } \frac{m n k}{m n+n k+k m}=\frac{1}{\frac{1}{m}+\frac{1}{n}+\frac{1}{k}} \\
=\frac{1}{3 x+3 y+3 z}=\frac{1}{3}.
\end{array}
$$