Here the gate equation is given as $A\oplus S=Q_{1}\oplus S$
Now XOR gate changes to XNOR gate and still the equation remain same, i.e. $A\odot S=Q_{1}\odot S$
Now both equation will be equal when $Q_{1}\oplus S=Q_{1}\odot S’=Q_{1}’\odot S$
So, in place of $Q_{1}$ if we place $Q_{1}’$ both will give similar output.