6. In $\triangle A B C$, $\angle A=30^{\circ}, \angle C=90^{\circ}$, with $C$ as the center and $C B$ as the radius, a circle is drawn intersecting side $A B$ at $M$ and side $A C$ at $N, C M$ intersects $B N$ at point $P$. If $A N=1$, then $S_{\triangle C P N}-S_{\triangle B P M}=(\quad)$.
(A) $\frac{1}{8}$
(B) $\frac{\sqrt{3}}{8}$
(C) $\frac{1}{4}$
(D) $\frac{\sqrt{3}}{4}$
6. (A).

As shown in the right figure, it is easy to know that $M$ is the midpoint of $AB$. Let $BC=x$, then we have $(2x)^{2}=x^{2}+(x+1)^{2}$, which simplifies to $x^{2}-x=\frac{1}{2}$. Therefore,
$$
\begin{array}{l}
S_{\triangle X P N}-S_{\triangle B P M}=S_{\triangle C B N}-S_{\triangle C R M} \\
=\frac{1}{2} x^{2}-\frac{1}{2} S_{\triangle A B C}=\frac{1}{2} x^{2}-\frac{1}{4} x(x+1) \\
=\frac{1}{4} \cdot\left(x^{2}-x\right)=\frac{1}{8} .
\end{array}
$$