6. Given that $P$ is the vertex of a circular cone whose axial section is an isosceles right triangle, $PA$ is a generatrix of the cone, and $B$ is a moving point on the base of the cone. Then the maximum value of $\frac{PA + AB}{PB}$ is

保留源文本的换行和格式，直接输出翻译结果。
6. $\sqrt{4+2 \sqrt{2}}$.

Let the center of the base circle be $O$, and the radius of the base be 1. Extend $A C$ to point $B'$ such that $O B = O B'$.
Thus, $P B = P B'$.
By $A B \leqslant A O + O B = A O + O B' = A B'$
$\Rightarrow \frac{P A + A B}{P B} \leqslant \frac{P A + A B'}{P B'}$.
The equality holds when $A$, $O$, and $B$ are collinear.
In $\triangle P A B'$, let $\angle A P B' = \alpha$, $\angle A B' P = \beta$.
Then $\alpha + \beta = 135^{\circ}$.
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{P A + A B'}{P B'} = \frac{\sin \beta + \sin \alpha}{\sin 45^{\circ}} \\
= \frac{2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2}}{\sin 45^{\circ}} \\
\leqslant \frac{2 \sin \frac{\alpha + \beta}{2}}{\sin 45^{\circ}} = \frac{2 \sin 67.5^{\circ}}{\sin 45^{\circ}} = \frac{1}{\sin 22.5^{\circ}} . \\
\text{When } \alpha = \beta, \\
\left(\frac{P A + A B'}{P B'}\right)_{\text{max}} = \frac{1}{\sin 22.5^{\circ}} = \sqrt{4 + 2 \sqrt{2}} .
\end{array}
$$