when is $\mathbb P_{\mathcal {A}}$ separative? This is exercise $II.16$ of Kunen's set theory. 
Some background: 

  
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*$\Bbb P_{\mathcal A}=\{\langle s,F\rangle:s\subseteq\omega, |s|<\omega, F\subseteq \mathcal A,|F|<\omega\}$, $\langle s,F\rangle\leq \langle s',F'\rangle$ iff $s'\subseteq s$,$F'\subseteq F$ and $\forall x\in F'(x\cap s\subseteq s')$
  
*Let $\Bbb P$ be a partial order, give $\Bbb P$ the topology of initial segments and let $\mathcal B$ be the set of all regular open sets of $\Bbb P$, then for all $p\in \Bbb P$ define $i(p)=\operatorname{int}(\operatorname{cl}(N_p))$.
  
*(Exercise $II.15$) $\Bbb P$ is separative iff $\Bbb P$ is an strict ordering and $$\forall p,q\in \Bbb P(p\leq q\leftrightarrow i(p)\subseteq i(q)).$$
  

I have proved that if $\mathcal A\subseteq P(\omega)$ and $i(\langle \Bbb s,F\rangle)\subseteq i(\langle \Bbb s',F'\rangle)$, then $\forall x\in F'(x\cap s\subseteq s')$, and $s'\subseteq s$ iff $\bigcup \mathcal A=\omega$, for all $\langle \Bbb s,F\rangle,\langle \Bbb s',F'\rangle\in \Bbb P$. But I haven't come up with a sufficient and necessary condition for $F'\subseteq F$ to hold.
So, is there such a condition?
Thanks
Edit: $\Bbb P$ is separative iff $\Bbb P$ is a strict order and whenever $p\nleq q$, there is $r\leq p$ such that $r\perp q$.
 A: Let $p=\langle s,F\rangle$ and $q=\langle t,G\rangle$, and suppose that $p\not\le q$. If $p\bot q$, we’re done, so assume that $p$ and $q$ are compatible. Then by Lemma $2.8$ 
$$\forall x\in G(x\cap s\subseteq t)\quad\text{and}\quad\forall x\in F(x\cap t\subseteq s)\;,\tag{1}$$
and $r_0=\langle s\cup t,F\cup G\rangle\in\Bbb P_{\mathcal A}$ with $r_0\le p,q$.
Now $p\not\le q$, so either $s\nsupseteq t$, $F\nsupseteq G$, or there is an $x\in G$ such that $x\cap s\nsubseteq t$, and $(1)$ rules out that last possibility.
Suppose that $s\nsupseteq t$, and fix $n\in t\setminus s$. We want to construct $r=\langle u,H\rangle\in\Bbb P_{\mathcal A}$ so that $r\le p$ and $r\bot q$. Let $u=s$; $H$ will be $F\cup\{x\}$ for a set $x\subseteq\omega$ yet to be determined, so that $r=\langle s,F\cup\{x\}\rangle$. This is already enough to ensure that $r\le p$. If we now require that $n\in x$, we have $x\cap t\nsubseteq s$, and it follows from Lemma $2.8$ that $r\bot q$.
Now suppose that $s\supseteq t$ and $F\nsupseteq G$. Fix $y\in G\setminus F$. If $y\setminus\left(s\cup\bigcup F\right)\ne\varnothing$, fix $n\in y\setminus\left(s\cup\bigcup F\right)$, and let $r=\langle s\cup\{n\},F\rangle$. Then $x\cap(s\cup\{n\})=x\cap s\subseteq s$ for each $x\in F$, so $r\le p$. But $$n\in\Big(y\cap(s\cup\{n\})\Big)\setminus s\subseteq\Big(y\cap(s\cup\{n\})\Big)\setminus t\;,$$ so $r\bot q$.
The simplest condition on $\mathcal A$ that I can think of to ensure that $y\setminus\left(s\cup\bigcup F\right)\ne\varnothing$ is to make $\mathcal{A}\subseteq[\omega]^\omega$ an almost disjoint family: if $y\in[\omega]^\omega$ and $y\subseteq s\cup\bigcup F$, then $y$ must have infinite intersection with some $x\in F$. Thus, $\Bbb P_{\mathcal A}$ is separative if $\mathcal{A}\subseteq[\omega]^\omega$ is almost disjoint.
