4 The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A$ and $B$. 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
Let a point on the ellipse be $P(4 \cos \alpha, 3 \sin \alpha)$. When it is on the opposite side of $A B$ from the origin $O$, the distance from $P$ to $A B$ is
$$
\begin{aligned}
& \frac{3(4 \cos \alpha)+4(3 \sin \alpha)-12}{5} \\
= & \frac{12}{5}(\cos \alpha+\sin \alpha-1) \leqslant \frac{12}{5}(\sqrt{2}-1) \\
< & \frac{6}{5},
\end{aligned}
$$

and $A B=5$, so, $S_{\triangle P A B}<\frac{1}{2} \times 5 \times \frac{6}{5}=3$.
Therefore, when the area of $\triangle P A B$ is 3, point $P$ is on the same side of $A B$ as $O$, and there are 2 such points, hence the answer is (B).