Let $\mathcal{F}$ be a $P$-filter on $\omega$. Denote by $\Omega=\bigsqcup \omega\_i$ where $\omega\_i=\omega$. Consider the $P$-filter $\mathcal{S}$ on $\Omega$ whose base is as follows
$(\bigsqcup\_i F\_i, F\_i\in \mathcal{F})$.


$\mathcal{F}$ filter is isomorphic to $\mathcal{S}$ filter ?



P-filter property?
After the clarifications in comments of OP (August 28 and today), here's a counterexample. Let $\mathcal F$ be the filter consisting of only $\omega-\{0\}$ and $\omega$. Then $\mathcal S$ (as defined in the question) contains a set whose complement is infinite, whereas $\mathcal F$ does not. So they cannot be isomorphic.


