4. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A, B$. There is a point $P$ on the ellipse such that the area of $\triangle P A B$ equals 3. How many such points $P$ are there?
A. 1
B. 2
C. 3
D. 4
4. For Figure B, considering the area $S$ of quadrilateral ${ }_{1} A O B$,
$$
\begin{aligned}
S= & S_{\triangle O P_{1}}+S_{\dot{\Delta} X w_{1}} \\
& =\frac{1}{2} \times 4(3 \sin \alpha)+\frac{1}{2} \times 3(4 \cos \alpha) \\
& =6(\sin \alpha+\cos \alpha) \\
& =6 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right),
\end{aligned}
$$

Therefore, $S_{\text {max }}=6 \sqrt{2}$ (when $\alpha=\frac{\pi}{4}$ ).
The area $S_{\triangle O A B}=\frac{1}{2} \times 4 \times 3=6$ is a constant,

Since $6 \sqrt{2}-6<3$,
point $P$ cannot be above the line $A B$, and there are clearly two points $P$ below the line $A B$, hence the answer is B.