This construction is a bit complicated and I am not sure it works. We claim that Alice can win. First, we will construct a single set such that procedure $P$ described in the problem is not periodic. Consider first $A=\{2^n\mid n\in\{0,2,4,\dots\}\}$. Then clearly $P(A)$ just contains the powers of $2$ which are not powers of $4$ and $P(P(A))=A$, etc. To make this non-periodic, we have to fill in other primes as follows: \[ \begin{matrix} 2\cdot 3 \\ &2^2\cdot 5\\ 2 \cdot 5 &&2^3\cdot 7\\ & 2^2\cdot 11 && 2^4\cdot 11 \end{matrix} \]i.e. for each prime we take every second power of $2$ (and we extend this "triangle" by one each row). Also, we add all numbers with exactly one prime factor to this (other than $2$ itself). Now clearly, the operation $P$ on this set will transform it $\{2,8,32,\dots\}$ and the numbers from the triangle as well, which will simply be the "gaps" that we skipped: \[ \begin{matrix} 2\cdot 5\\ &2^2\cdot 7\\ 2\cdot 11 &&2^3\cdot 11 \end{matrix} \]etc. Now clearly we lose a row of this triangle at each step, so the set we get never repeats. To obtain a partition from this, we repeat the exact same trick for powers of $3$, i.e. we take all powers of $3$ which are not powers of $9$ and not previously taken. To this, we add all numbers with at most two prime factors and build a triangle as above \[ \begin{matrix} 3^2\cdot 5\\ &3^3\cdot 7\\ 3^2\cdot 11 & & 3^4\cdot 11 \end{matrix} \]etc. The same argument as above shows that iterating $P$ on this set never repeats. We can build a partition of all positive integers this way, at each step using the $k$-th prime, taking the union with such a triangle and adding all numbers with at most $k$ prime factors. Clearly, this forms a partition of $\mathbb{Z}^+$. Also, note that the numbers with at most $k$ prime factors "die out" after sufficiently many steps, so this is also not a problem. Finally, we have to show that we cannot arrive at the same set by starting with different $A,B\in\mathcal{C}$ and applying $P$. For this, assume $A,B$ were created in the above procedure with primes $p,q$. Then clearly $A$ and all of $P(A),P(P(A)),\dots$ will contain numbers with arbitrarily high $\nu_p$, and similarly $B$ contains numbers with arbitrarily high $\nu_q$. In addition, if $A$ was constructed in the $k$-th step, then all primes $\neq p$ divide elements of $A$ at most $k$-times. Thus, $A$ and all of $P(A),P(P(A)),\dots$ cannot be equal to any one of $B,P(B),\dots$.