Vertices of quadrilaterals as circumcenters

Such points $P$, $Q$, $R$, $S$ do not always exist.

To see this, consider the (arguably) simplest quadrilateral - the unit square. We place this in coordinate space with $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$.

If $A$ is the circumcenter of $\triangle QRS$, then $A$ is equidistant from $Q$, $R$, and $S$. Similarly, $B$ is equidistant from $R$, $S$, and $P$. Since $A$, $B$ are both equidistant from $R$, $S$, we have that $AB$ is the perpendicular bisector of $RS$. Hence, $x_R=x_S$ and $y_R+y_S=0$.

Similarly, we have $x_S+x_P=2$, $y_P+y_Q=2$, and $x_Q+x_R=0$.

Define $x_R=x$. Then, $x_P=2-x$, $x_Q=-x$, and $x_S=x$. Also, define $y_R=y$. This forces $y_S=-y$.

Since $B$ is equidistant from $R$ and $P$, $$(x-1)^2+y^2=(1-x)^2+y_P^2 \implies y_P= \pm y$$

Since $C$ is equidistant from $S$ and $P$, $$(x-1)^2+(-y-1)^2=(x_P-1)^2+(y_P-1)^2$$ Since $x_P=2-x$, $$(x-1)^2+(-y-1)^2=(1-x)^2+(y_P-1)^2 \implies y_P=-y \text{ or } y_P=y+2.$$

So, either $y=0$ (degenerate case since $R$ and $S$ would coincide) or $y_P=-y$.

Since $D$ is equidistant from $R$ and $Q$, $$x^2+(y-1)^2=x_Q^2+(y_Q-1)^2$$ Since $x_Q=-x$, $$(y-1)^2=(y_Q-1)^2 \implies y_Q=y \text{ or } 2-y.$$

However, we have from earlier that $y_P+y_Q=2$. Since $y_P=-y$, $y_Q=2+y$ regardless. But then the only possibility is for $y=0$, which we already established to lead to a degenerate case.