If (p∧∼r)⇒(q ∧ r) is false and q and r are both false, then p is
True
p⇒q is false only when p is true and q is false.
Given that (p ∧∼r)⇒(q ∧ r) is false.
Hence, p ∧∼r is true and q ∧ r is false.
Given that q and r are both false. Hence, q ∧ r is false.
Since r is false, ∼r is true.
p ∧∼r must be true. Hence, p must be true.