4. 16 Detailed Explanation: The equations of the asymptotes of the hyperbola are $y= \pm \frac{4}{3} x$. Let $P_{1}\left(x_{1}, \frac{4}{3} x_{1}\right), P_{2}\left(x_{2},-\frac{4}{3} x_{2}\right), P\left(x_{0}, y_{0}\right)$. Given $\lambda=\frac{\overrightarrow{P_{1} P}}{\overrightarrow{P P_{2}}}=3$, we get $P\left(\frac{x_{1}+3 x_{2}}{4}, \frac{x_{1}-3 x_{2}}{3}\right)$. Substituting into the hyperbola equation and simplifying, we get $x_{1} x_{2}=12$. Using the angle formula, we get $\tan \angle P_{1} O P_{2}=\frac{24}{7}$, thus $\sin \angle P_{1} O P_{2}=\frac{24}{25}$. Using the triangle area formula, we get
$$
S=\frac{1}{2}\left|O P_{1}\right|\left|O P_{2}\right| \sin \angle P_{1} O P_{2}=\frac{1}{2} \cdot \frac{25}{9} \cdot\left|x_{1} x_{2}\right| \sin \angle P_{1} O P_{2}=16
$$
4. Let point $P$ be a point on the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. A line passing through $P$ intersects the two asymptotes at $P_{1}$ and $P_{2}$, and $\overrightarrow{P_{1} P} \overrightarrow{P P_{2}}$ $=3$. Let $O$ be the origin. Then the area of $\triangle O P_{1} P_{2}$ is $\qquad$.