Thai Mathematical Olympiad

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

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

From the power of the point $M$, $P{M}^2=M{A}^2=ME\cdot MC$.

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

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 $△PAN\sim △POA$, $\frac{PA}{PN}=\frac{PO}{PA}$. Thus, $$PN\cdot PO=P{A}^2=PD\cdot PC.$$ 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.