The statement  ~(p∨q)∨(~p∧q) is equivalent to

Use De-Morgan's and distributive law.We have, ~(p∨q)∨(~p∧q)≡(~p∧~q)∨(~p∧q)[ by De Morgan's law ∼(p∨q)=(∼p∧∼q)]≡∼p∧(~q∨q)  [by distributive law]≡∼p∧t       [~q∨q=t]≡∼p