If p and q are logical statements, then p ⇒ ( q ⇒ p) is equivalent...

$p ⇒ \left( q ⇒ p\right) ≡ ∼ p ∨ \left(q ∨ p\right)$
$≡ \left(∼ p ∨ p\right) ∨ \left(q\right) ≡ t ∨ \left(q\right) ≡ t$
Hence, the given statement $p ⇒ \left( q ⇒ p\right)$ is a tautology
Now, $p ⇒ \left(p ⇒ q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology
$p ⇒ \left(q ∨ p\right) ≡ p ∨ \left(p ∨ q\right) ≡ \left( p ∨ p\right) ∨ q$
$t ∨ q ≡ t$
Hence, $p ⇒ \left(q ∨ p\right)$ is a tautology.
Now, $p ⇒ \left(p ∧ q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology
Also, $p ⇒ \left(p ⇔ q\right)$ is false when $p$ is true and $q$ is false

Hence, not a tautology