Given: $(\alpha: (\forall x)\left[P_x \Leftrightarrow (\forall y)\left[Q_{xy} \Leftrightarrow \neg Q_{yy} \right]\right] \Rightarrow (\forall x)\left[\neg P_x\right])\\$
$Q_{yy}$ is always True, this makes $\neg Q_{yy}$ False.
Writing $(\forall y)[Q_{xy} \Leftrightarrow \neg Q_{yy}]$ is same as writing $(\forall y)[Q_{xy} \Leftrightarrow False]]$
This is equivalent to saying that, for all y $Q_{xy}$ is false and finally we can rewrite $(\forall y)[Q_{xy} \Leftrightarrow \neg Q_{yy}]]$ as $(\forall y)[\neg Q_{xy}]$
α: (∀x)[P_{x}⇔(∀y)[¬Q_{xy}]]⇒(∀x)[¬P_{x}]
LHS:(∀x)[P_{x}⇔(∀y)[¬Q_{xy}]]
consider only (∀y)[¬Q_{xy}] it says all values of y does not divide x, but there will be atleast one value of y (when y=x, or when y=1) that divides x, i.e. [¬Q_{xy}] is not true for all values of y. (∀y)[¬Q_{xy}] is false.
Now LHS becomes (∀x)[P_{x}⇔False], "P_{x}⇔False" this means P_{x }is False, which is same as writing "¬P_{x}"_{.}
Finally, we reduced LHS to (∀x)[¬P_{x}]
α: (∀x)[¬P_{x}]⇒(∀x)[¬P_{x}] (which is trivial, P(x)⇒ P(x) is trivially true)
Hence (∀x)[¬P_{x}]⇒(∀x)[¬P_{x}] is trivially true for any P(x), doent matter if P(x) is for prime number or for composite number (I1 or I2).
⇒ I1 and I2 both satisfies α.
Option D.
Thanks Soumya29 for suggesting edit.