We show two solutions.

Solution to proposed problem We let $\overline{PP_2}$ and $\overline{P_1P_3}$ intersect (perpendicularly) at point $K_2$, and define $K_\bullet$ cyclically.

Claim — The points $K_\bullet$ are concyclic say with circumcircle $\gamma$.

Proof. Note that $PP_1 \times PK_1 = PP_2 \times PK_2 = \dots$ so the result follows by inversion at $P$. $\square$

Let $E_i$ be the second intersection of line $\overline{P_{i-1}K_iP_{i+1}}$ with $\gamma$; then it follows that the perpendiculars to $\overline{P_{i-1}P_{i+1}}$ at $E_i$ all concur at a point $E$, which is the reflection of $P$ across the center of $\gamma$.

We let $H_2 = \overline{P_1P_3} \cap \overline{P_2P_4}$ denote the orthocenter of $\triangle PP_2P_3$ and define $H_\bullet$ cyclically.

Claim — We have
$$\overline{EH_2} \perp \overline{P_1P_4} \parallel \overline{K_2K_3} \text{ and } \overline{PH_2} \perp \overline{E_2E_3} \parallel \overline{P_2P_3}.$$
Proof. Both parallelisms follow by Reim's theorem through $\angle E_2H_2E_3 = \angle K_2H_2K_3$, So we need to show the perpendicularities.
Note that $\overline{H_2P}$ and $\overline{H_2E}$ are respectively circum-diameters of $\triangle H_2K_2K_3$ and $\triangle H_2E_2E_3$.
As $\overline{K_2K_3}$ and $\overline{E_2E_3}$ are anti-parallel, it follows $\overline{H_2P}$ and $\overline{H_2E}$ are isogonal and we derive both perpendicularities. $\square$

Claim — The points $E, Q_3, E_3$ are collinear.

Proof. We use the previous claim. The parallelisms imply that
$$\frac{E_3 H_2}{E_3 P_2} = \frac{E_2 H_2}{E_2 P_3} = \frac{E_4 H_3}{E_4 P_3} = \frac{E_3 H_3}{E_3 P_4}.$$
Now consider a homothety centered at $E_3$ sending $H_2$ to $P_2$ and $H_3$ to $P_4$. Then it should send the orthocenter of $\triangle EH_2H_3$ to $Q_3$, proving the result. $\square$

From all this it follows that $\triangle EQ_2Q_3 \sim \triangle PK_2K_3$ as the opposite sides are all parallel.
Repeating this we actually find a homothety of 100-gons
$$Q_1 Q_2 Q_3 \cdots \sim K_1 K_2 K_3 \cdots$$
and that concludes the proof.

Solution to generalization (Nikolai Beluhov) We are going to need some well-known lemmas.

Lemma
Suppose that $ABCD$ is inscribed in a circle $\Gamma$. Let the tangents to $\Gamma$ at $A$ and $B$ meet at $E$, let the tangents to $\Gamma$ at $C$ and $D$ meet at $F$, and let diagonals $AC$ and $BD$ meet at $P$. Then points $E, F$, and $P$ are collinear.

Proof. Let the circle of center E and radius EA = EB meet lines AC and BD for the second time at points U and V. By a simple angle chase, triangles EUV and FCD are homothetic. $\Box$

Lemma
Suppose that points X and Y are isogonal conjugates in polygon $A = A_1A_2...A_n$. (This means that lines $A_iX$ and $A_iY$ are symmetric with respect to the interior angle bisector of $\angle A_{i-1}A_iA_{i+1}$ for all $i$, where $A_{n+j} \equiv A_j$ for all $j$.) Then the 2n projections of X and Y on the sides of $A$ are concyclic.

Proof. By a simple angle chase, for all $i$ we have that the four projections on sides $A_{i-1}A_i$ and $A_iA_{i+1}$ are concyclic. Say that they lie on circle $\Gamma_i$. Consider the midpoint $M$ of segment $XY$. For every side $s$ of $A$, we have that $M$ is equidistant from the projections of $X$ and $Y$ on $s$. Therefore, $M$ is the center of $\Gamma_i$ for all $i$, and thus all of the $\Gamma_i$ coincide. $\Box$

Lemma
Let $\Gamma'$ and $\Gamma''$ be two circles and let $r$ be some fixed real number. Then the locus of points $X$ such that $\text{Pow}(X, \Gamma')$: $\text{Pow}(X, \Gamma'') = r$ is concyclic.

Proof. This is a classical result in circle geometry. A full proof is given, for example, in item 115 of Roger Johnson's Advanced Euclidean Geometry. $\Box$

We are ready to solve the problem. Let $\mathcal{P}$ be our polygon, let $O$ be its the circumcenter, and let $\Gamma$ be its circumcircle.
Fix any index $i$. In triangle $P_{i-1}P_iP_{i+1}$, we have that line $P_iP$ contains the altitude through $P_i$ and line $P_iO$ contains the circumradius through $P_i$. Therefore, these two lines are symmetric with respect to the interior angle bisector of $\angle P_{i-1}P_iP_{i+1}$.
Thus points $P$ and $O$ are isogonal conjugates in $\mathcal{P}$. By Lemma 2, it follows that the projections of $O$ onto the sides of $\mathcal{P}$ are concyclic. In other words, the midpoints of the sides of $\mathcal{P}$ lie on some circle $\omega$.

Let $M_i$ be the midpoint of segment $P_iP_{i+1}$ and let the tangents to $\Gamma$ at points $P_i$ and $P_{i+1}$ meet at $T_i$. Since inversion with respect to $\Gamma$ swaps $M_i$ and $T_i$ for all $i$, and also since all of the $M_i$ lie on the same circle $\omega$, we obtain that all of the $T_i$ lie on the same circle $\Omega$.

Again, fix any index $i$. By Lemma 1 applied to cyclic quadrilateral $P_{i-2}P_{i-1}P_{i+1}P_{i+2}$, we have that point $Q_i$ lies on line $T_{i-2}T_{i+1}$. Similarly, point $Q_{i+1}$ lies on line $T_{i-1}T_{i+2}$.

Define
$$f_i = \frac{\mathrm{Pow}(Q_i, \Gamma)}{\mathrm{Pow}(Q_i, \Omega)}.$$

Claim — We have $f_i = f_{i+1}$ for all $i$.

Proof. Note that
$$\begin{aligned} \mathrm{Pow}(Q_i, \Gamma) &= Q_i P_{i-1} \cdot Q_i P_{i+2} \\ \mathrm{Pow}(Q_{i+1}, \Gamma) &= Q_{i+1} P_{i-1} \cdot Q_{i+1} P_{i+2} \\ \mathrm{Pow}(Q_i, \Omega) &= Q_i T_{i-2} \cdot Q_i T_{i+1} \\ \mathrm{Pow}(Q_{i+1}, \Omega) &= Q_{i+1} T_{i-1} \cdot Q_{i+1} T_{i+2}. \end{aligned}$$

Consider cyclic quadrilateral $T_{i-2}T_{i-1}T_{i+1}T_{i+2}$. Since $\Gamma$ touches its opposite sides $T_{i-2}T_{i-1}$ and $T_{i+1}T_{i+2}$ at points $P_{i-1}$ and $P_{i+2}$, we have that line $P_{i-1}P_{i+2}$ makes equal angles with these opposite sides. From here, a simple angle chase shows that triangles $P_{i-1}Q_iT_{i-2}$ and $P_{i+2}Q_{i+1}T_{i+2}$ are similar. Thus
$$\frac{Q_i P_{i-1}}{Q_i T_{i-2}} = \frac{Q_{i+1} P_{i+2}}{Q_{i+1} T_{i+2}}.$$
Similarly,
$$\frac{Q_i P_{i+2}}{Q_i T_{i+1}} = \frac{Q_{i+1} P_{i-1}}{Q_{i+1} T_{i-1}}.$$
From these, the desired identity $f_i = f_{i+1}$ follows. $\square$

Therefore, the power ratio $f_i$ is the same for all $i$. By Lemma 3 for circles $\Gamma$ and $\Omega$, the solution is complete.