Let ⊕ and ⊗ are two mathematical operators. If p ⊕ (p ⊗ q) is not a...

\left(A\right)p ∨ \left(p ⇒ q\right) is false only if $pis false and \left(p ⇒ q\right) is false which is never possible simultaneously. Hence,$p ∨ \left(p ⇒ q\right)$is a tautology<br /> \left( C\right)p ⇒ \left(p ∨ q\right) is false only if$pis true and \left( p ∨ q\right) is false which is never possible simultaneously. Hence,$p ⇒ \left(p ∨ q\right)$ is a tautology
\left(B\right)p ⇒ \left(p ∧ q\right) is false only if $p$ is true & $p ∧ q$ is false which is possible when $p$ is true & $q$ is false. Hence, $p ⇒ \left(p ∧ q\right)$ is not a tautology