It's all about the angles.
Repeated use of isosoles triangles gives:
$\displaystyle m\angle LSD + 2m\angle LDS = 180$, and $\displaystyle m\angle ASD + 2m\angle SAD = 180$
$\displaystyle m\angle ASB + 2m\angle SAB = 180$, and $\displaystyle m\angle BSK + 2m\angle SBK = 180$
Thus:
$\displaystyle (m\angle ASD + m\angle LSD) + 2(m\angle LDS + m\angle SAD) = 360$
$\displaystyle (m\angle ASB + m\angle BSK) + 2(m\angle SAB + m\angle SBK) = 360$
From parallelogram m$\displaystyle \angle LDA = m\angle ABK$, so
$\displaystyle m\angle LDS + m\angle SDA = m\angle ABS + m\angle SBK$
But via isosoles, $\displaystyle m\angle SDA = m\angle SAD$, and $\displaystyle m\angle ABS = m\angle SAB$.
Thus
$\displaystyle m\angle ASD + m\angle LSD$
$\displaystyle = 360 - 2(m\angle LDS + m\angle SAD)$
$\displaystyle = 360 - 2(m\angle LDS + m\angle SDA)$
$\displaystyle = 360 - 2(m\angle ABS + m\angle SBK)$
$\displaystyle = 360 - 2(m\angle SAB + m\angle SBK)$
$\displaystyle = m\angle ASB + m\angle BSK$.
Now, using that $\displaystyle \overline{AN}$ is a diameter, get
$\displaystyle m\angle ASD + m\angle LSD+ m\angle NSL = 180$
$\displaystyle m\angle ASB + m\angle BSK+ m\angle NSK = 180$
Thus
$\displaystyle m\angle NSL = 180 - (m\angle ASD + m\angle LSD)$
$\displaystyle = 180 - (m\angle ASB + m\angle BSK)$
$\displaystyle = m\angle NSK$.
But from $\displaystyle m\angle NSL = m\angle NSK$ and Side-Angle-Side, get
$\displaystyle \triangle KSN \cong \triangle LSN$, and so $\displaystyle m\overline{NL} = m\overline{NK}$.