Problem 433

4. (A).

It is easy to know that $\triangle A M E \backsim \triangle A B F$, thus,
$$S_{\triangle A M E}=\left(\frac{A E}{A F}\right)^{2} S_{\triangle A B F}=\frac{1}{5} S_{\triangle A B F}$$
Since $N$ is the centroid of $\triangle A B C$,
$\therefore S_{\triangle B F N}=\frac{1}{3} S_{\triangle A B F}$.
Since $S_{\triangle D E B}=S_{\text {A.1PY }}$.
$$\begin{aligned} \therefore S_{\triangle A K N} & =S_{\triangle A M K}-S_{\triangle B F N}=\frac{8}{15} S_{\triangle A B F} \\ & =\frac{8}{15} \times \frac{1}{2} \times 2 \sqrt{15} \times \frac{2 \sqrt{15}}{2}=8 . \end{aligned}$$