6. As shown in the figure, $A B C D$ is a square inscribed in $\odot O$, and $P$ is any point on $\overparen{A D}$. Then the value of $\frac{P A+P C}{P B}$ is ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2
6. (B).

Draw a perpendicular from point $B$ to $BP$ intersecting the extension of $PC$ at point $E$.
In $\triangle B A P$ and $\triangle B C E$,
$$
\begin{array}{l}
\angle B C E=180^{\circ}-\angle B C P \\
=\angle B A P, \\
\angle C B E=90^{\circ}-\angle P B C \\
=\angle A B P, \\
B C=A B, \\
\therefore \triangle B C E \cong \triangle B A P. \\
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } C E=P A. \\
\therefore(P A+P C): P B=(C E+P C): P B \\
\quad=P E: P B=\sqrt{2} .
\end{array}
$$