The statement p → (q → p) is equivalent to
p → (p → q)
p → (p ∨ q)
p → (p ∧ q)
B.
p → (q → p) = ~ p ∨ (q → p) = ~ p ∨ (~ q ∨ p) since p ∨ ~ p is always true = ~ p ∨ p ∨ q = p → (p ∨ q).