7. In quadrilateral $A B C D$, $A B=B C, \angle A=$ $\angle C=90^{\circ}, \angle B=13 E^{\prime}, K$ is a point on $A B$, $N$ is a point on $B C$. If the perimeter of $\triangle B K N$ is equal to 2 times $A B$, then the degree measure of $\angle K D N$ is
7. Connect $B D$, obviously $\mathrm{Rt} \triangle A B D \cong \mathrm{Rt} \triangle G B D$. Extend $B C$ to $A_{1}$, such that $C A_{1}=A K$, it is easy to prove $\mathrm{Rt} \triangle D C A_{1} \cong \mathrm{Rt} \triangle D A K$.
Therefore, $\angle A_{1} D K=\angle C D A=\angle 45^{\circ}$.
Also, $K N+B K+B N=2 A B=A B+B C$
$$
\begin{array}{l}
=B K+K A+B N+N C \\
=B K+B N+(N C+K A),
\end{array}
$$

then $K N=N A$. Thus, $\triangle K D N \cong A_{1} D N$.
Hence, $\angle K D N=22.5^{\circ}$.