We can consider all given points as complex numbers, points of the complex plane. Then, as far as I understood, for all integers $m\ge 1$, $1\le j\le k$ we we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2},$$
where $\xi=\exp\frac{2\pi i}{k}$.
Since $A_{m+1,1}= A_{m,3}$, we have
$$A_{m+1}+(P_{m+1}-A_{m+1})\xi^{-1}=A_m+(P_m-A_m)\xi.$$
It follows
$$A_{m+1}=-A_m\xi+\frac{P_m\xi^2-P_{m+1}}{\xi-1}.$$
This formula suggests that the chain can be closed for a specific choice of $O$ and $P_m$’s, but I guess that you are looking for a stable general pattern. So let’s look when the sequence $A_n$ is periodic.
Putting $B_m=A_m(-\xi)^m$, we obtain
$$B_{m+1}=B_m+\frac{P_m\xi^2-P_{m+1}}{\xi-1}(-\xi)^{m-1}.$$
Since the sequence $\{P_n\}$ has period $n$, we have
$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^n,$$
that is
$$A_{m+2n}-A_{m+n}(1+\xi^{-n})+A_m=0.$$
The following cases are possible.
1)) $\xi^{-n}=1$, that is $k|n$. Then $A_{m+2n}-2A_{m+n}+A_m=0$, that is a sequence $\{A_{m+rn}\}_{r\in\Bbb N}$ is an arithmetic progression. It is periodic iff $A_m=A_{m+n}$. Since $k|n$, we have
$$0=A_{m+n}(-\xi)^{m+n}- A_{m}(-\xi)^{m}=B_{m+n}-B_{m}=$$ $$\frac {(-\xi)^{m-1}}{\xi-1}\sum_{j=1}^n - P_{m+j} (-\xi)^{j-1}(\xi^3+1)=$$ $$-\frac {(-\xi)^{m-1}(\xi^3+1)}{\xi-1}\sum_{j=1}^n P_{m+j} (-\xi)^{j-1}.$$
$$\sum_{j=1}^n P_{m+j} (-\xi)^{j-1}=0,$$
that is, $-\xi$ is a root of a polynomial $P(x)=\sum_{j=1}^n P_{j} x^j$. It follows that $A_{m+n}=A_m$ for each $m$, that is, the the sequence $\{A_m\}$ has period $n$.
2)) $\xi^{-n}\ne 1$, that is $k\not|n$. It is easy to check that an equation $\lambda^2-(1+\xi^{-n})\lambda +1$ has two distinct roots $\lambda_1,\lambda_2$. The theory of recurrence relations implies that $A_{r n+m}=C_1\lambda_1^r+ C_2\lambda_2^r$ for each $r$ and some constants $C_1$ and $C_2$. Since $\lambda_1\lambda_2=1$, if not $|\lambda_1|=|\lambda_2|$ and the sequence $A_{rn+m}$ is non-zero then $|A_{rn+m}|$ either grows to infinity or vanishes to zero. It both cases the sequence $A_{rn+m}$ is not periodic. If $|\lambda_1|=|\lambda_2|$ then, since $\lambda_1\lambda_2=1$, we have $\lambda_2=\overline{\lambda_1}$, and so $1+\xi^{-n}=-(\lambda_1+\lambda_2)$ is a real number. Thus $\xi^{-n}$ is a real number distinct from $1$, that is $\xi^{-n}=-1$, that is $n\equiv k/2 \pmod k$ . Then $\lambda_{1,2}=\pm 1$, and so $A_{rn+m}=C_1+ C_2\cdot (-1)^r$ for each $r$ and some constants $C_1$ and $C_2$. It follows $A_m=A_{m+2n}$. Since $k|2n$, we have
$$0=A_{m+2n}(-\xi)^{m+2n}- A_{m}(-\xi)^{m}=B_{m+n}-B_{m}=$$ $$\frac {(-\xi)^{m-1}}{\xi-1}\sum_{j=1}^{2n} - P_{m+j} (-\xi)^{j-1}(\xi^3+1)=$$ $$-\frac {(-\xi)^{m-1}(\xi^3+1)}{\xi-1}\sum_{j=1}^{2n} P_{m+j} (-\xi)^{j-1}.$$
$$\sum_{j=1}^{2n} P_{m+j} (-\xi)^{j-1}=0,$$ that is, $-\xi$ is a root of a polynomial $(1+x^n)P(x)=\sum_{j=1}^{2n} P_{j} x^j$. It follows that $A_{m+2n}=A_m$ for each $m$, that is, the sequence $\{A_m\}$ has period $n$.