The correct option is D neither a fallacy nor a tautology
The truth table of given expression is given below
pq∼p∼qx≡(p ∨∼q)y≡(∼p∨q)x∧yTTFFTTTTFFTTFFFTTFFTFFFTTTTT
Clearly, the truth value of given preposition is sometimes true and sometimes false, hence (p ∨∼q)∧(∼p∨q) is neither a fallacy nor a tautology.