Well, a quadrilateral inside a quadrilateral

$OM \text{is angle bisector} \implies \angle BOM = \angle AOM \quad \cdots Eq. 1 \\ OQ \text{ is angle bisector} \implies \angle DOQ = \angle AOQ \quad \cdots Eq . 2 \\ \text{adding both equation we get, } \\ \implies \angle BOM + \angle DOQ = \angle AOM + \angle AOQ = \angle MOQ \\ \implies \angle BOM + \angle DOQ + \angle MOQ = 2\angle MOQ \\ \implies 2\angle MOQ = 180^{\circ} \\ \implies \angle MOQ = 90^{\circ}. \\ \implies MP \bot NQ \\ \\ \text{Diagonals of quadrilateral } MNPQ \text{ intersect at } 90^{\circ}. \quad \cdots \textbf{Statement 1}$

$\Delta AOD \text{ and } \Delta COB \text{ are congruent } \Bigg[ \because AD = CB , AO = OC \text{ and } DO = OB \Bigg] \\ \implies OQ = ON \quad\bigg[ \because Q \text{ and } N \text{ are corresponding in-centres of the triangles }\bigg] \\ \text{ Similarly } OM = OP \\ \text{ Diagonals of quadrilateral } MNPQ \text{ bisect each other. } \cdots \textbf{Statement 2}$

$\text{Combining both the statements, } \\ \textbf{Diagonals of quadrilateral } MNPQ \textbf{ bisect each other at } 90^{\circ}. \\ \text{ and this is the definition of } \textbf{rhombus}. \\ \implies MNPQ \text{ is a } \textbf{rhombus}.$