6. Given points $P$, $Q$, $R$ are on the sides $AB$, $BC$, $CA$ of $\triangle ABC$ respectively, and $\quad BP=PQ=QR=RC=1$.
Then the maximum area of $\triangle ABC$ is ( ).
(A) $\sqrt{3}$
(B) 2
(C) $\sqrt{5}$
(D) 3
6. B.

Notice that
$S_{\triangle B P Q} \leqslant \frac{1}{2} \times 1 \times 1 \times \sin \angle B P Q \leqslant \frac{1}{2}$.
Similarly, $S_{\triangle P Q R} \leqslant \frac{1}{2}, S_{\triangle Q R C} \leqslant \frac{1}{2}$.
Let $\triangle D P R$ be the reflection of $\triangle Q P R$ about $P R$.
If point $D$ coincides with $A$, then
$S_{\triangle P A R}=S_{\triangle P Q R} \leqslant \frac{1}{2}$.
If point $D$ does not coincide with $A$, then
$$
\begin{array}{l}
\angle D=\angle P Q R \\
=180^{\circ}-(\angle B+\angle C)=\angle A .
\end{array}
$$

Therefore, points $A, P, R, D$ are concyclic.
It is easy to see that the distance from point $A$ to line $P R$ is less than the distance from point $D$ to line $P R$, hence $S_{\triangle P A R}<S_{\triangle P Q R} \leqslant \frac{1}{2}$.
Thus, $S_{\triangle A B C} \leqslant 2$.
When $\angle A=90^{\circ}, A B=A C=2$, the equality holds.