8. Given that the volume of tetrahedron $P-ABC$ is $1$, $G$ and $K$ are the centroids of $\triangle ABC$ and $\triangle PBC$ respectively. A line through $G$ intersects $AB$ and $AC$ at points $M$ and $N$. Then the maximum volume of the quadrilateral pyramid $K-MNCB$ is
8. $\frac{5}{27}$.

In $\triangle ABC$, let
$$
\overrightarrow{AM}=x \overrightarrow{AB}, \overrightarrow{AN}=y \overrightarrow{AC},
$$
$D$ is the midpoint of $BC$.
Then $\overrightarrow{AG}=\frac{2}{3} \overrightarrow{AD}=\frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})$.
Thus, $\overrightarrow{MG}=\overrightarrow{AG}-\overrightarrow{AM}=\left(\frac{1}{3}-x\right) \overrightarrow{AB}+\frac{1}{3} \overrightarrow{AC}$,
$\overrightarrow{GN}=\overrightarrow{AN}-\overrightarrow{AG}=-\frac{1}{3} \overrightarrow{AB}+\left(y-\frac{1}{3}\right) \overrightarrow{AC}$.
Since $\overrightarrow{MG}=\lambda \overrightarrow{GN}$, we have
$$
\frac{\frac{1}{3}-x}{-\frac{1}{3}}=\frac{\frac{1}{3}}{y-\frac{1}{3}} \Rightarrow \frac{1}{x}+\frac{1}{y}=3.
$$

Using the AM-GM inequality, we get $xy \geqslant \frac{4}{9}$.
Therefore, $S_{\triangle AMN} \geqslant \frac{4}{9} S_{\triangle ABC}, S_{\text{quadrilateral } MNCB} \leqslant \frac{5}{9} S_{\triangle ABC}$.
Since $K$ is the centroid of $\triangle PBC$, then