Let $X=\{a,b,c\}$ and $\mathcal{P}X=\{\emptyset,X,\{a\},\{b\},\{c\},\{a,b\}.\{a,c\},\{b,c\}\}$.
I can only see 4 subalgebras of $\mathcal{P}X$, namely:
$\mathcal{F}_0=\{\emptyset,X\}$
$\mathcal{F}_1=\{\emptyset,X,\{a\},\{b,c\}\}$
$\mathcal{F}_2=\{\emptyset,X,\{b\},\{a,c\}\}$
$\mathcal{F}_1=\{\emptyset,X,\{c\},\{a,b\}\}$
Is there any rule to determine the number of subalgebras of a power set algebra one can form?
For a finite set $X$ of cardinality $n$, the boolean subalgebras of $\mathcal P X$ correspond to the equivalence relations on $X$, i.e. partitions of $X$ into "atoms". The number of these (including $\mathcal P X$ itself, which you didn't include in your four for $n=3$) is the Bell number $B_n$.