Let $AN \cap BL = P$.

a) Since $\angle LAM = \angle MAN = \angle LBM = \angle MBN$, the quadrilaterals $ABED$ and $ABFG$ are cyclic. Then $\angle AED = \angle ABL = \angle ANL$ and hence $DE\parallel LN$. Analogously $FG\parallel LN$.

b) Since $\frac{DP}{LP} = \frac{DE}{LN} = \frac{GF}{LN} = \frac{CF}{CN}$, then $\frac{LD}{DP} = \frac{NF}{FC}$. Hence
$$\frac{LA}{AP} = \frac{LD}{DP} = \frac{NF}{FC} = \frac{NA}{AC}.$$
It follows that $\triangle APL \sim \triangle ACN$ which gives $\angle APL = \angle ACB$, i.e. $LPNC$ is a cyclic quadrilateral. Then
$$\begin{align*} 180^\circ &= \angle APB + \angle ACB = 180^\circ - \angle PAB - \angle PBA + 180^\circ - \angle CAB - \angle CBA \\ &= 2 \cdot 180^\circ - (\angle PAB + \angle CAB) - (\angle PBA + \angle CBA) \\ &= 2(180^\circ - \angle MAB - \angle MBA) = 2 \angle AMB, \end{align*}$$
i.e. $\angle AMB = 90^\circ$. So $DF \perp EG$ and therefore the parallelogram $DEFG$ is a rhombus.