The statement (p ⇔ q) is not equivalent to

Option $\left(A\right)$ is theorem hence \left(A\right)$is equivalent to \left(p ⇔ q\right)$
Option \left( B\right) \left( p ⇔ q\right) means only $p$ or only $q$
Which is same as \left(p ∧ q\right) ∨ \left( q ∧ p\right)
Option \left(C\right) \left( p ∨ q\right) ∧ \left( pv q\right)
$≡ \left( p ∨ q\right)$ ∧ \left( \left(p ∧ q\right)\right)
Means \left(p ∨ q\right)$happens but \left(p ∧ q\right)$ does not happen. It means only $p$ or only $q$ happens
Option \left(D\right)$\left(p ⇔ q\right)$ ≡ \left( p ∨ q\right)∨ \left( \left( p ∧ q\right)\right)$<br />$≡ \left( \left( p ∨ q\right) ⇒ \left( p ∧ q\right)\right)<br />but not equivalent to \left(p ∨ q\right) ⇒ \left( p ∧ q\right)