If the Boolean expression ( p ∧ q ) ⊛ ( p ⊗ q ) is a tautology,...

Option $\left(A\right)$

$( p ∧ q ) \to ( p \to q )$

We know that $a \to b = ~ a ∨ b$

$= ( p ∧ q ) ⟶ ( ~ p ∨ q )$

$= ~ ( p ∧ q ) ∨ ( ~ p ∨ q )$

Using De Morgan's Law i.e. $~ \left(a ∧ b\right) = ~ a ∨ ~ b,$

$= ( ~ p ∨ ~ q ) ∨ ( ~ p ∨ q )$

Using associative law and commutative law i.e. \left(a ∨ b\right) ∨ c = a ∨ \left(b ∨ c\right) \text{ and } a ∨ b = b ∨ a,$</p><p>$= ~ p ∨ ( ~ q ∨ q )$</p><p>$= ~ p ∨ t$</p><p>$= t$</p><p>Option$( B )$</p><p>$( p ∧ q ) ∧ ( p ∨ q ) = ( p ∧ q )$</p><p>Which is not a tautology.</p><p>Option$( C )$</p><p>$( p ∧ q ) ∨ ( p \to q )$</p><p>$= ( p ∧ q ) ∨ ( ~ p ∨ q )$</p><p>$= ~ p ∨ q$</p><p>Which is not a tautology.</p><p>Option$( D )$</p><p>$( p ∧ q ) ∧ ( p \to q )$</p><p>$= ( p ∧ q ) ∧ ( ~ p ∨ q )$</p><p>$= p ∧ q$</p><p>Which is not a tautology.</p><p>Hence, the correct answer is option \left(A\right) .</p>$