Construct a truth table for (p ^ q) ↔ ~p
the truth table for bi-conditional <-> is
it is true when both p & q are true, or both are false.
p q (p^q) <-> ~p
T T T T T F F
T F T F F F T
F T F F T F F
F F F F F F T
Since in all cases, the proposition is false, it is a contradiction.