Four. (18 points) As shown in Figure 6, given $\triangle A B C$, and $S_{\triangle A B C}=$ 1. $D, E$ are moving points on $A C, A B$ respectively, $B D$ and $C E$ intersect at point $P$, such that $S_{\text {quadrilateral } B C D E}=$ $\frac{16}{9} S_{\triangle B P C}$. Find the maximum value of $S_{\triangle D E P}$.
Let $\frac{A E}{A B}=x, \frac{A D}{A C}=y$, then
$\frac{S_{\triangle A E D}}{S_{\triangle A B C}}=\frac{A E \cdot A D}{A B \cdot A C}=x y$.
Since $S_{\triangle A B C}=1$, we have
$S_{\triangle A D E}=x y, S_{\text {quadrilateral } C D E}=1-x y$.
In $\triangle A B D$, by Menelaus' theorem, we get
$$
\frac{B P}{P D} \cdot \frac{D C}{C A} \cdot \frac{A E}{E B}=1 \text {. }
$$

Thus, $\frac{B P}{P D}=\frac{C A}{D C} \cdot \frac{E B}{A E}=\frac{1}{1-y} \cdot \frac{1-x}{x}=\frac{1-x}{x(1-y)}$.
Similarly, $\frac{C P}{P E}=\frac{1-y}{y(1-x)}$.
Therefore, $\frac{S_{\triangle P P}}{S_{\triangle B E C}}=\frac{P D \cdot P E}{P B \cdot P C}=\frac{x(1-y)}{1-x} \cdot \frac{y(1-x)}{1-y}=x y$.
Also, from equation (1), we have
$$
\frac{B P}{B D}=\frac{B P}{B P+P D}=\frac{1-x}{1-x y} \text {. }
$$

Thus, $S_{\triangle B A C}=\frac{B P}{B D} S_{\triangle B C D}=\frac{1-x}{1-x y} \cdot \frac{C D}{C A} S_{\triangle \triangle B C}$
$$
=\frac{(1-x)(1-y)}{1-x y} \text {. }
$$

By the problem statement,
$$
\begin{array}{l}
\frac{(1-x)(1-y)}{1-x y}=S_{\triangle B R C} \\
=\frac{9}{16} S_{\text {quadrilateral BCDE }}=\frac{9}{16}(1-x y) .
\end{array}
$$

Let $u=x y$, then
$$
9(1-u)^{2}=16[1+u-(x+y)] \leqslant 16(1-\sqrt{u})^{2} \text {. }
$$

Noting that $0<u<1$, we get
$$
3(1-u) \leqslant 4(1-\sqrt{u}), 3(1+\sqrt{u}) \leqslant 4 \text {, }
$$

Solving, we get $0<u \leqslant \frac{1}{9}$.
When and only when $x=y=\frac{1}{3}$, $u=x y=\frac{1}{9}$.
Then $S_{\triangle A B P}=x y S_{\triangle B R C}=x y \frac{9}{16}(1-x y)=\frac{9}{16}\left(-u^{2}+u\right)$
$$
=\frac{9}{16}\left[-\left(u-\frac{1}{2}\right)^{2}+\frac{1}{4}\right] \text {. }
$$

Therefore, when $u=\frac{1}{9}$, i.e., $x=y=\frac{1}{3}$, $S_{\triangle \text { esp }}$ reaches its maximum value $\frac{1}{18}$.