Let p, q, and r be propositions and the expression (p → q) → r be a contradiction. Then the expression (r → p) → q is

(p → q) → r is a CONTRADICTION which is possible only when r is FALSE and (p → q) is true.Now, from here we can clearly SAY that option 4 is correct as (r → p) → q means ¬ (r → p) ∨ q.Since r is false, (r → p) is true and ¬ (r → p) becomes false.So, it becomes (false ∨ q). Now it totally DEPENDS on q. Whenever q is true, this value will always be true.Alternate Method:Let X ≡ p → q, Y ≡ (p → q) → r ≡ X → r,Z ≡ r → p and W ≡ (r → p) → q ≡ Z → pUsing TRUTH tablepqrXYZW00010100011101010101101111011000 110101011011010111111111 So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.