p⇒(∼q⇒p)≡simp∨(q∨p) ≡(simp∨p)∨(q)≡t∨(q)≡t
Hence, the given statement p⇒(∼q⇒p) is a tautology
Now, p⇒(p⇒q) is false when p is true and q is false
Hence, not a tautology p⇒(q∨p)≡∼p∨(p∨q)≡(∼p∨p)∨q t∨q≡t
Hence, p⇒(q∨p) is a tautology.
Now, p⇒(p∧q) is false when p is true and q is false
Hence, not a tautology
Also, p⇒(p⇔q) is false when p is true and q is false
Hence, not a tautology