Observe that $AB \perp NP$. Thus, $M$ is the circumcenter of $\triangle ANP$ and hence $MN = MP$.

It can also be seen that $MN \parallel PQ$.

From the power of the point $M$, $PM^2 = MA^2 = ME \cdot MC$.

So, $\frac{PM}{ME} = \frac{MC}{PM}$ and hence $\triangle PME \sim \triangle CMP$.

Thus, $M\hat{P}E = M\hat{C}P$.

Since $O$, $A$, $P$, $B$ are cyclic, the power of the point $N$ tells us that
$$CN \cdot NE = AN \cdot NB = ON \cdot NP.$$
Thus, $C$, $P$, $E$, $O$ are also cyclic and hence $E\hat{P}N = N\hat{C}O$.

Since $\triangle PAN \sim \triangle POA$, $\frac{PA}{PN} = \frac{PO}{PA}$. Thus,
$$PN \cdot PO = PA^2 = PD \cdot PC.$$
So, $C$, $D$, $N$, $O$ are cyclic and hence $Q\hat{N}P = P\hat{C}O$.

We can now see that
$$Q\hat{N}P = P\hat{C}O = P\hat{C}M + M\hat{C}O = M\hat{P}E + E\hat{P}N = M\hat{P}N.$$
Thus, $MP \parallel NQ$. Therefore, $MNQP$ is a rhombus.