@Arjun Sir plz verify this method.
$Q_{yy}$ 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, It is true irrespective of x being prime or composite. Even if x is composite then only its factor divides x not all values.
Now LHS becomes (∀x)[P_{x}⇔True], "P_{x}⇔True" this means P_{x }is True, which is same as writing "P_{x}" only_{.}
Finally, we reduced LHS to (∀x)[P_{x}]
α: (∀x)[P_{x}]⇒(∀x)[¬P_{x}]
LHS: (∀x)[P_{x}].
LHS is false for I_{1} bcoz not all x are primes False ⇒ Anything is True. For I_{1} α is true hence I_{1} satisfies α.
Similarly, LHS is false for I_{2} also bcoz not all x are composite. hence I_{2} satisfies α.
Option D.