Let $P$ and $M$ be marked on $AD$ and $BC$ so that $MP \perp AD$ and $BM = MC$. By condition, $K$ belongs to $MP$. Let $L$ and $N$ be respectively the feet of the perpendiculars from $K$ to the lines $AB$ and $DC$. Two cases are possible:

1) Both the points lie on the sides $AB$ and $DC$ (or on their extensions).

2) One of the points lies on the side and the other point lies on the extension of the other side.

Since $AK$ and $DK$ are bisectors, we have $KL = KP = KN$. Since $K$ lies on the perpendicular bisector, we have $BK = KC$. Therefore the right-angled triangles $BLK$ and $CNK$ are equal, so $\angle LBK = \angle NCK$, $BL = CN$.

For the first case (see Fig. 1) from the isosceles triangle $BKC$ it follows that $\angle KBC = \angle KCB$. So
$$\begin{aligned} \angle ABC &= \angle ABK + \angle KBC = (180^\circ - \angle LBK) + \angle KBC = \\ &= (180^\circ - \angle NCK) + \angle KCB = \angle KCD + \angle KCB = \angle BCD. \end{aligned}$$
Therefore, $\angle ABC = \angle DCB$, and so the trapezium $ABCD$ is isosceles. (Similarly we can consider the cases when both $L$ and $N$ lie on the sides $AB$ and $DC$).

Fig. 1

If we have the second case (see Fig. 2), then $AL = AP$, $PD = DN$ (since $AK$ and $DK$ are bisectors). Therefore,
$$AB + DC = (AL + LB) + (DN - NC) = AP + PD + (LB - NC) = AP + PD = AD,$$
as required.

Fig. 2