6. As shown in Figure 2, in rectangle $A B C D$, it is known that $A B=4$, $B C=2$. Points $F$ and $E$ are on $A B$ and $C D$ respectively. Connecting $D F$, $C F$, $A E$, and $B E$ intersect at points $Q$ and $P$. If the area of quadrilateral $P E Q F$ is $S$, then ( ).
(A) $S \geqslant 2$
(B) $S \leqslant 2$
(C) $S \geqslant 4$
(D) $S \leqslant 4$
6. B.

As shown in Figure 5(a), in trapezoid $ABCD$, $AC \parallel BD$, and $AD$ intersects $BC$ at point $E$. Let
\[
\begin{array}{l}
S_{\triangle AEC}=a, S_{\triangle AEB}=c, \\
S_{\triangle BED}=b, S_{\triangle DEC}=d .
\end{array}
\]

Then $ab=cd$, and $c=d$.
Therefore, $c=d=\sqrt{ab}$.
Also, $(\sqrt{a}-\sqrt{b})^{2} \geqslant 0$, so
\[
a+b \geqslant 2 \sqrt{ab}=c+d \text{. }
\]

Thus, from Figure 5(b), we know
\[
\begin{array}{l}
2(c+f) \leqslant a+b+d+e . \\
\text{Hence } 4(c+f) \leqslant a+b+d+e+2c+2f=8 .
\end{array}
\]

Therefore, $S=c+f \leqslant 2$.