6. As shown in Figure 2, given
$$
\begin{array}{l}
\angle D A P=\angle P B C= \\
\angle C D P=90^{\circ}, A P= \\
P B=4, A D=3 . \text { Then } \\
B C=(\quad) .
\end{array}
$$
(A) $\frac{32}{3}$
(B) 16
(C) $\frac{41}{3}$
(D) $\frac{41}{2}$
6. C.

Connect $P C$, and draw $D E \perp B C$ at point $E$.
It is easy to know that $P D=5, D E=8$.
Let $E C=x$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
P C^{2}=P D^{2}+D C^{2}=25+64+x^{2}, \\
P C^{2}=P B^{2}+B C^{2}=16+(3+x)^{2} \\
\Rightarrow 89=25+6 x \Rightarrow x=\frac{32}{3} .
\end{array}
$$

Thus, $B C=3+\frac{32}{3}=\frac{41}{3}$.