Solution:

Let $AC \cap BD = O$ and $X$ be the second intersection point of the circumcircles of $\triangle AOB$ and $\triangle BOC$. Set $\text{XBO = XCO =}$ and $\text{XAO = XDO =}$. Since $\triangle AXC \sim \triangle DXB$, then $\frac{XD}{XA} = \frac{BD}{AC}$. This and the condition of the problem implies that $\frac{AP}{AC} = 1 - \frac{BQ}{BD} = \frac{DQ}{BD}$ and hence $\frac{AP}{DQ} = \frac{AC}{BD} = \frac{XA}{XD}$. Then $\triangle APX \sim \triangle DQX$ and so $\text{APX = DQX}$. The last equality means that the points $X, Q, O$ and $P$ are concyclic and thus $\text{XQP = XOP = XDA}$.

This implies that the points $X, Q, D$ and $M$ are concyclic, i.e. the circumcircle of $\triangle DMQ$ passes through $X$. Then $\text{XMN =}$ and hence the points $X, A, P$ and $M$ are concyclic, i.e., the circumcircle of $\triangle AMP$ passes through $X$.

It follows in the same way that the circumcircle of $\triangle CNP$ passes through $X$ and then analogously the circumcircle of $\triangle BNQ$ passes through $X$.