Such an easy problem Let $ N \equiv QB \cap AD.$ Then $ QM$ is the Q-median of $ \triangle QBC.$ Since $ BC \parallel ND,$ then $ P$ is midpoint of $ ND.$ The triangle $ \triangle NDK$ is isosceles. So if $ R \equiv KD \cap BC,$ then $ \triangle KBR$ is also isosceles $ \Longrightarrow$ $ \angle QBC = \angle BRK = \angle KDA.$

Let $B' = QB%Error. "capAD" is a bad command. $ and $R = KD%Error. "capBC" is a bad command. $. Notice that we only need to prove that $\triangle KB'D$ is isosceles, because if this happens, $\triangle KBM$ is also isosceles(because $B'D // BC$), and so $\angle KBR=\angle QBC = \angle KRB = \angle CRD=\angle RDA=\angle KDA$. But $BC // B'D$, and so there're an homothety that goes from $\triangle QBC$ to $\triangle QND$, and if $M$ is the midpoint of $BC$, $P$ is the midpoint of $B'D$, what implies that $\triangle KND$ is isosceles. Q.E.D.

Angle $ADK$=Angle $DMC$=Angle $KXM$=Angle $KBM$. This is just one situation. $X$ is the point of intersection between $DK$ and $BC$.

From Luis's solution we get $KP$ external bisector of angle $\angle QKD$, hence the parallel to the bases of trapezoid through $K$ is the internal angle bisector of $\angle QKD$, done. Best regards, sunken rock