[asy][asy] import olympiad; size(8cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair O = origin, A = (-1,1), B = (1,1), C = (1,-1), D = (-1,-1), M = (0.25,-1), X1 = 2*foot(dir(270),O,M)-dir(270), Q = extension(B,C,M,X1), P = 2*foot(C,O,M)-C, N = extension(C,D,O,foot(O,D,P)), R = extension(P,N,A,D), K = (M+N)/2, S = (C+D)/2, T = foot(Q,A,D); draw(arc(O,1,180,360)); draw(arc(O,sqrt(2),225,315)); draw((1,0)--C--D--(-1,0)); draw(K--O^^R--Q, linewidth(0.8)); draw(R--P--Q); draw(Q--T^^O--S, dashed); dot("$C$", C, dir(330)); dot("$D$", D, dir(210)); dot("$P$", P, dir(250)); dot("$O$", O, dir(90)); dot("$Q$", Q, dir(0)); dot("$R$", R, dir(180)); dot("$T$", T, dir(180)); dot("$S$", S, dir(270)); dot("$K$", K, dir(270)); dot("$N$", N, dir(240)); dot("$M$", M, dir(285)); [/asy][/asy] WLOG assume $CP>DP$. Since the incircle and circumcircle of $ABCD$ are concentric, we must have $ND = NP$ and $MC = MP$ by symmetry. It follows that $$\triangle RND \cong \triangle MNP \cong \triangle MQC.$$Let $S$ be the projection of $O$ to $\overline{CD}$, $T$ be the projection of $Q$ to $\overline{AD}$, $r$ be the radius of the incircle, and $K$ be the midpoint of $\overline{MN}$. Then we have $$\frac{QT}{OS} = \frac{2r}{r} = 2$$and $$\frac{RT}{KS} = \frac{RD-QC}{\tfrac 12 (CK-DK)} = \frac{PM-PN}{\tfrac 12 (CM - DN)} = \frac{2(PM-PN)}{PM-PN} = 2,$$implying $\triangle QTR \sim \triangle OSK$ with ratio $2$, so $OK = \tfrac 12 QR$. Also $$\angle(\overline{OK}, \overline{QR}) = \angle(\overline{OS}, \overline{QT}) + \angle KOS - \angle RQT = 90^\circ,$$which solves the other part.

Let $P_2$, $P_3$, $P_4$ be points on the circumcircle such that $PP_2P_3P_4$ forms a square. For convenience, let's change $P$ to $P_1$. As $P_1P_2P_3P_4$ is a rotation of $ABCD$ around $O$, they have the same incircle. Let the other intersection points of the squares be $X, Y, U, V$. $MNRXYUVQ$ forms a nice octagon. Let $K, L, S, T$ be midpoints of the segments $MN, YU, RX, QV$ respectively. $RXVQ$ is a parallelogram, hence $ST\parallel QR\parallel XV$. The octagon contains a bunch of rotations of $90^{\circ}$ one of which is $ST$ and $KL$, so $ST\perp KL$. Also, $2OK=KL=ST=QR$. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); import olympiad; pair A = dir(135), B = dir(45), C = dir(315), D = dir(225); real angle = 73; pair O = (0, 0), P1 = dir(180 + angle), P2 = dir(90 + angle), P3 = dir(angle), P4 = dir(angle - 90); pair M = extension(P1, P4, C, D), N = extension(P1, P2, C, D); pair Q = extension(P1, M, B, C), R = extension(P1, N, A, D); pair X = extension(P3, P2, A, D), Y = extension(P3, P2, A, B); pair U = extension(P3, P4, A, B), V = extension(P3, P4, B, C); pair S = midpoint(X -- R), T = midpoint(V -- Q), K = midpoint(M -- N), L = midpoint(Y -- U); draw(unitcircle); draw(circle(O, 1/sqrt(2))); draw(A -- B -- C -- D -- cycle); draw(P1 -- P2 -- P3 -- P4 -- cycle, dashed); draw(M -- N -- R -- X -- Y -- U -- V -- Q -- cycle, linewidth(1.2)); draw(X -- O -- Q ^^ N -- O -- U, blue); draw(Y -- O -- M ^^ R -- O -- V, red); draw(Q -- R ^^ X -- V, gray); draw(S -- T ^^ K -- L, gray); dot("$A$", A, dir(135)); dot("$B$", B, dir(45)); dot("$C$", C, dir(315)); dot("$D$", D, dir(225)); dot("$P_1$", P1, dir(180 + angle)); dot("$P_2$", P2, dir(90 + angle)); dot("$P_3$", P3, dir(angle)); dot("$P_4$", P4, dir(angle - 90)); dot("$O$", O, dir(90)); dot("$M$", M, dir(270)); dot("$N$", N, dir(270)); dot("$Q$", Q, dir(0)); dot("$R$", R, dir(180)); dot("$X$", X, dir(180)); dot("$Y$", Y, dir(90)); dot("$U$", U, dir(90)); dot("$V$", V, dir(0)); dot("$S$", S, dir(180)); dot("$T$", T, dir(0)); dot("$K$", K, dir(270)); dot("$L$", L, dir(90)); [/asy][/asy]