By the given condition clearly $P_kQP_{k+4}$ collinear for all $k$. Now so $QP_k.QP_{k+4}=l$ for all $k$ hence that sum is greater equals to $8l$ so equality iff $QP_k=QP_{k+4}$ for all $k$ and hence clearly $Q$ is center.

Claim. Let $A,B,C,D$ be concyclic and let $AC$ and $BD$ intersect at $E$ such that $\angle AEB=45^\circ$. Then, $AB^2+CD^2\geq 2(2-\sqrt{2})r^2$. Proof. Note that $\angle AOB+\angle COD=90^\circ$. By Law of Cosines, $$AB^2=2r^2-2r^2\cos{\angle AOB}$$and $$CD^2=2r^2-2r^2\cos{\angle COD},$$thus we need $$\sqrt{2}\geq \cos{\angle AOB}+\cos{\angle COD}.$$Note that $\cos$ is concave on $[0^\circ,90^\circ]$, thus by Jensen's inequality, the claim follows. [asy][asy]import olympiad; size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,E; O=(0,0);A=dir(60);B=dir(100);C=dir(300);D=dir(330);E=extension(A,C,B,D); draw(A--B--C--D--cycle,deep);draw(circumcircle(A,B,C));draw(A--C,med);draw(B--D,med); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$O$",O,S); [/asy][/asy] Hence by the claim, $$\sum_{i=1}^{8}P_{i - 1} P_i^2 \geq 8(2-\sqrt{2})r^2.$$Note that equality holds iff $\angle P_iOP_{i+1}=45^\circ$, which means that $P_1\ldots P_8$ is regular, hence $Q$ is the center of the circle.