II. The side length of square $ABCD$ is 1. With vertex $C$ as the center and $CB$ as the radius, draw a circle. On $AB$, take $BP = \frac{1}{5}$. Draw a tangent line from $P$ to the circle, intersecting $AD$ at $Q$, with the point of tangency being $E$. Connect $CE$ to intersect the diagonal $BD$ at $K$. Connect $AK$ to intersect $PQ$ at $M$. Find the length of $AM$.
Given that points $C, D, Q, E$ are concyclic, then $\angle 3=\angle 1$.
Given that points $C, E, P, B$ are concyclic, then $\angle 4=\angle 2$.
$$
\begin{array}{l}
\because \frac{D K}{K B}=\frac{S_{\triangle D C K}}{S_{\triangle K C B}} \\
=\frac{\sin \angle 1}{\sin \angle 2}=\frac{\sin \angle 3}{\sin \angle 4} . \\
\frac{D K}{K B}=\frac{S_{\triangle D A K}}{S_{\triangle K A B}}=\frac{\sin \angle 5}{\sin \angle 6}, \\
\therefore \frac{\sin \angle 3}{\sin \angle 4}=\frac{\sin \angle 5}{\sin \angle 5} . \\
\frac{Q M}{M P}=\frac{A(2 \sin }{\angle 3}=\frac{\sin \angle 4}{A P \sin \angle 6} \cdot \frac{\sin \angle 5}{\sin \angle 3}=1 .
\end{array}
$$

Therefore, $M$ is the midpoint of $P Q$.
Let $D \dot{Q}=x, A Q=1-x$. Since $B P=\frac{1}{5}, A P=\frac{4}{5}$, $P Q=\frac{1}{5}+x$ (tangent), then
$$
(1-x)^{2}+\left(\frac{4}{5}\right)^{2}=\left(\frac{1}{5}+x\right)^{2} \text {. }
$$

Solving for $x$ gives $x=\frac{2}{3}$. Hence $P Q=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$.
Therefore, $A M=\frac{1}{2} P Q=\frac{13}{30}$.