Let $a$ be the rhombus' side length. Then one can find $u, v, t \in (0, 1)$ such that $AM = au$, $BN = av$, and $DP = at$.

Denote by $G$ the centroid of $MNP$ and suppose that the diagonals of the rhombus intersect at $O$. Then $G \in AC$ if and only if there exists some $k \in \mathbb{R}$ such that $\overline{OG} = k\overline{OA}$.

On the other hand,
$$\begin{align*} 3\overline{OG} &= \overline{OM} + \overline{ON} + \overline{OP} \\ &= u\overline{OB} + (1-u)\overline{OA} + v\overline{OC} + (1-v)\overline{OB} + t\overline{OC} + (1-t)\overline{OD} \\ &= (1-u-v-t)\overline{OA} + (u+1-v-1+t)\overline{OB} \\ &= (1-u-v-t)\overline{OA} + (u-v+t)\overline{OB}. \end{align*}$$
We conclude that $G \in AC$ if and only if $u - v + t = 0$, which is equivalent to $AM + DP = BN$.