4. As shown in Figure 2, the graphs of two inverse proportion functions $y=\frac{k_{1}}{x}$ and $y=\frac{k_{2}}{x}\left(k_{1}>k_{2}>\right.$
$0)$ in the first quadrant are curves $C_{1}$
and $C_{2}$, respectively. Let point $P$ be on $C_{1}$,
$P E \perp x$-axis at point $E$, intersecting $C_{2}$ at point $A$, and $P D \perp y$-axis at point $D$, intersecting $C_{2}$ at point $B$. Then the area of quadrilateral $P A O B$ is $(\quad)$.
(A) $k_{1}+k_{2}$
(B) $k_{1}-k_{2}$
(C) $k_{1} k_{2}$
(D) $\frac{k_{1}}{k_{2}}$
4. B.

Let $P(m, n)$. Since point $P$ is on $C_{1}$, we have,
$$
\begin{array}{l}
m n=k_{1}, A\left(m, \frac{k_{2}}{m}\right), B\left(\frac{k_{2}}{n}, n\right) . \\
S_{\text {quadrilateral } P A O B}=S_{\text {quadrilateral } P E O D}-S_{\triangle O E 1}-S_{\triangle O B D} \\
=m n-\frac{1}{2}\left(m \cdot \frac{k_{2}}{m}+n \cdot \frac{k_{2}}{n}\right)=k_{1}-k_{2} .
\end{array}
$$