all logic concept is based on $T\rightarrow F$ .
1st let's see for $I_{2}$
If we somehow make a combination such that the implication will false, then we're done.
Now we have only one way to make an implication false, which is $T\rightarrow F$ combination.
If the domain is composite(means not prime) then right side of the $\alpha$ is always T, which implies we can't make $\alpha$ false. $P_{x} = F$ i.e $\sim P_{x} = T$ always.
So, $I_{2}$ satisfies $\alpha$.
Now, let's come in $I_{1}$
In $I_{1}$, $\sim {Q_{yy}}$ is always false because in any circumstances y divides y.
now, try to make $T\rightarrow F$ combination.
let's forget about quantifiers & treat $\alpha$ as $[P_{x} \Leftrightarrow [Q_{xy}\Leftrightarrow Q_{yy}]] \rightarrow \sim P_{x}$.
take $P_{13}$, RHS is F. LHS is $[P_{13} \Leftrightarrow [Q_{13.1}\Leftrightarrow Q_{1.1}]]$
So, $[P_{13} \Leftrightarrow [T\Leftrightarrow F]]$
$[P_{13} \Leftrightarrow F]$ = $F$, which means $F\rightarrow F$ implies T
So, using any number(composite or prime), we can't make the implication false.
So, $I_{1}$ also satisfies $\alpha$.
Option D is the answer.