Using the properties of associativity and commutativity, and the below mentioned properties we can find the correct answer:
1. $X⊕X = 0$
2. $X⊕1=X'$
3. $X⊕0=X$
$F(P,Q)=((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0))$
$=(P'⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0))$
$=((P'⊕P)⊕Q))⊕((P⊕Q)⊕(Q⊕0))$
$=((1⊕Q)⊕((P⊕Q)⊕(Q⊕0))$
$=(Q'⊕((P⊕Q)⊕(Q⊕0))) $
$=(Q'⊕((P⊕Q)⊕Q)) $
$=(Q'⊕(P⊕(Q⊕Q))) $
$=(Q'⊕(P⊕0)) $
$=(Q'⊕P) $
$=(Q⊕P)'$
So, the correct option is, option no. D.