The statement ~ [p ∨ (~ (p ∧ q))] is equivalent to

Given,

$~ \left[p ∨ \left(~ \left(p ∧ q\right)\right)\right]$

Now we know that,

By demorgan law, $~ \left(A ∨ B\right) ≡ ~ A ∧ ~ B$

Now using above formula we get,

$~ \left[p ∨ \left(~ \left(p ∧ q\right)\right)\right]$

$≡ ~ p ∧ \left(p ∧ q\right)\left[\text{as} ~ \left(~ A\right) ≡ A\right]$

≡ \left(p ∧ q\right) ∧ \left(~ p\right)