By Pappus Theorem we have that $M,N$ and $S=AC\cap BD$ are collinear. From here it is obvious $BL=DK$ since $\bigtriangleup DKS \cong \bigtriangleup SLB$ (by $ASA$)

Extend $KL$ and let it meet $AB$ at $X$. Menalaus theorem on triangles $ADQ$,$APQ$,$QCB$ gives $$\frac {DM}{MQ} \frac{QX}{AX} \frac {AK}{KD}=1$$. $$\frac {DN}{NB} \frac {BX}{AX} \frac {AM}{MP}=1$$$$\frac {CL}{LB} \frac {BX}{QX} \frac {QN}{NC}=1$$and for parallel sides $$\frac {AM}{MQ}=\frac {DM}{MQ}$$and $$\frac {QN}{NC}=\frac {NB}{PN}$$solving and some coputation, I think we will get the result

Let $AC \cap BD = O$, clearly $BO = DO$. If it turns out that $O$ lies on the line $MN$, then it would follow that $\triangle DKO \cong \triangle BLO$ by side and two angles and hence $BL = DK$. Choose coordinates so that $A(0,0)$, $B(1,0)$ and $C(a,b)$ - then $D(a-1,b)$, as the segments $AB$ and $CD$ are equal and parallel. Let $Q(q,0)$ and $P(p,b)$. We may compute $M$ and $N$ via the equations of the lines $AP$, $DQ$, $BP$, $CQ$, but let us be smarter. By Thales' theorem we have $\frac{AM}{MP} = \frac{AQ}{DP} = \frac{q}{p-a+1}$, so \[M\left(\frac{0+\frac{q}{p-a+1}p}{1+\frac{q}{p-a+1}},\frac{0+\frac{q}{p-a+1}b}{1+\frac{q}{p-a+1}}\right) = M\left(\frac{pq}{p+q-a+1},\frac{bq}{p+q-a+1}\right).\]Analogously $\frac{QN}{NC} = \frac{BQ}{CP} = \frac{1-q}{a-p}$, whence \[ N\left(\frac{q+\frac{1-q}{a-p}a}{1 + \frac{1-q}{a-p}}, \frac{0 + \frac{1-q}{a-p}b}{1+\frac{1-q}{a-p}}\right) = N\left(\frac{a-pq}{a-p-q+1},\frac{(1-q)b}{a-p-q+1}\right).\]Hence as $O\left(\frac{a}{2},\frac{b}{2}\right)$, the slopes of the lines $OM$ and $ON$ are \[ \frac{\frac{bq}{p+q-a+1} - \frac{b}{2}}{\frac{pq}{p+q-a+1} - \frac{a}{2}} = \frac{b (a - p + q - 1)}{a^2 - a(p + q + 1) + 2 p q}, \ \ \frac{\frac{(1-q)b}{a-p-q+1} - \frac{b}{2}}{\frac{a-pq}{a-p-q+1} - \frac{a}{2}} = \frac{b (a - p + q - 1)}{a^2 - a (p + q + 1) + 2 p q} \]i.e. they are equal and since the lines have a common point (namely $O$), they coincide. This completes the proof.