(p′∨q′)∧(p′⊕q′) = p′⊕q′ holds since the satisfying assignments of p′⊕q′ are contained in the satisfying assignments of p′∨q′ (which has the additional one where both p′ and q′ are true). Hence, the intersection of both is just the set of satisfying assignments of p′⊕q′

For, (!p′∨q′)∧(p′⊕q′) = (p'?0:q'), we made a case distinction, i.e., Shannon decomposition on p': If p' is true, then we have (!1∨q′)∧(1⊕q′) which is q′∧(1⊕q′), i.e., q′∧!q′, thus 0. If p' is false, we get (!0∨q′)∧(0⊕q′) which is (1∨q′)∧(0⊕q′) = 1∧q′ = q′.