In fact, the statement is still true if we replace Assume that there exists a rectangle $ABCD$ and a point $P$ in the plane such that $PA,PB,PC,PD$ form an arithmetic progression in some order. By the British flag theorem, $PA^2+PC^2=PD^2+PB^2$, so WLOG the order of the progression is $PA,PB,PD,PC$. Then $PC-PD=PB-PA$, but $PB^2-PA^2=PC^2-PD^2$, so $PB+PA=PC+PD$. Then $PC=PB$ and $PD=PA$, contradicting the condition that the progression is strictly increasing.

For the sake of contradiction, assume that there exists such rectangle $ABCD$. Note that $ABCD$ is cyclic. Let $\omega$ be the smallest circle, with radius $a$. Let other circles have radii $a+d$, $a+2d$ and $a+3d$. Define $f(\bullet,\omega,(ABCD))=P(\bullet, \omega)-P(\bullet, (ABCD))$. Thus, $f$ is linear. Note that $f(A)=0$, $f(B)=(a+d)^2-a^2=2ad+d^2$, $f(C)=(a+2d)^2-a^2=4ad+4d^2$ and $f(D)=(a+3d)^2-a^2=6ad+9d^2$. But, note that $f(D)>f(B)+f(C)$. However as $ABCD$ is rectangle, its midpoints of diagonals must coincide, which means that $f(X)+f(Y)=f(Z)+f(W)$, where $(X,Y,Z,W)$ is a permutation of $(A,B,C,D)$. This yields desired contradiction. [asy][asy] size(9cm);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; O=(0,0);path a=circle(O,1);path b=circle(O,4/3);path c=circle(O,5/3);path d=circle(O,6/3);A=dir(100);B=4/3*dir(150);C=5/3*dir(0);path k=circumcircle(A,B,C);D=intersectionpoints(d,k)[0]; draw(a,blue);draw(b,green);draw(c,red);draw(d,orange);draw(k,purple); dot("$O$",O,S); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,S);dot("$D$",D,S); [/asy][/asy]