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 $\{A_m\}$ 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 a period $n$, we have 

$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^{-n},$$

that is 

$$A_{m+2n}-A_{m+n}(1+(-\xi)^{n})+A_m(-\xi)^{n}=0.$$


An equation $\lambda^2-(1+(-\xi)^{n})\lambda +(-\xi)^{n}$ has roots $1$ and $(-\xi)^{n}$. The following cases are possible. 

1))  $(-\xi)^{-n}=1$. This holds iff ($n$ is even and $k|n$) or ($n$ is odd, $k$ is even and $k|2n$). The theory of recurrence relations implies that $A_{r n+m}=c_1(m) + c_2(m)r$ for each $r$ and some constants $c_1(m)$ and $c_2(m)$ depending on $m$. If all $c_2(m)$ are zeroes then the sequence $\{A_m\}$ has a period $n$ (or its divisor). Otherwise the sequence $\{A_m\}$ is not periodic. Thus the sequence $\{A_m\}$ is periodic iff for each $m$ we have $A_m=A_{m+n}$. This can happen iff the choice of $P_m$’s is specific. Namely, 

1.1)) If $n$ is even and $k|n$ then 
 
$$0=A_{m+n}(-\xi)^{-m-n}- A_{m}(-\xi)^{-m}=B_{m+n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^n  P_{m+j} (-\xi)^{-(m+j)},$$

that is, $(-\xi)^{-1}$ is a root of a polynomial $P(x)=\sum_{j=1}^n  P_{j} x^j$. 

1.2)) If $n$ is odd, $k$ is even, and $k|2n$ then
 
$$0=A_{m+2n}(-\xi)^{-m-2n}- A_{m}(-\xi)^{-m}=B_{m+2n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^{2n}  P_{m+j} (-\xi)^{-(m+j)},$$

that is, $(-\xi)^{-1}$ is a root of a polynomial $(1+x^n)P(x)=\sum_{j=1}^{2n}  P_{j} x^j$

2)) $(-\xi)^{-n}\ne 1$. (This case holds, in particular, when both $n$ and $k$ are odd). The theory of recurrence relations implies that $A_{r n+m}=c_1(m) + c_2(m)(-\xi)^{nr}$ for each $r$ and some constants $c_1(m)$ and $c_2(m)$ depending on $m$. If all $c_2(m)$ are zeroes then the sequence $\{A_m\}$ has a period $n$ (or its divisor). Otherwise $-\xi$ is a primitive $q$-th root of unity, where 
$$q=\cases{k, \mbox{ if }k\equiv 0\pmod 4\\
k/2, \mbox{ if }k\equiv 2\pmod 4\\
2k, \mbox{ if }k\equiv 1,3\pmod 4}.$$

Remark that $q=\frac{2k}{\gcd(k-2,2k)}=\frac{2k}{\gcd(k-2,4)}$. Thus $(-\xi)^n$ is a primitive $\tfrac{q}{\gcd(q,n)}$-th root of unity, and so the sequence $\{A_m\}$ has a period $\tfrac{qn}{\gcd(q,n)}=\operatorname{lcm}(q,n)$ (or its divisor). Moreover, for each $m$, points $\{A_{r n+m}: 0\le r\le q-1\}$ are vertices of a $q$-qon. 

Finally, recall that for each $m\ge 1$, $1\le j\le k$ we have $A_{m,j}=A_m+(P_m-A_m)\xi^{j-2}$. It follows that if the sequence $\{A_m\}$ has a period $p$ then for each fixed $j$ a sequence $\{A_{m,j}\}$ has a period $\operatorname{lcm}(p,n)$ (or its divisor).