We need to simplify the above expression. As the given operation is XOR, we shall see property of XOR. Let A and B be boolean variable. In A XOR B, the result is 1 if both the bits/inputs are different, else 0. Now,
( ( 1 X P) X (P X Q) ) X ( (P X Q) X (Q X 0) )
( P' X P X Q ) X ( P X Q X Q ) ( as 1 X P = P' and Q X 0 = Q )
(1 X Q) X ( P X 0) ( as P' X P = 1 , and Q X Q = 0 )
Q' X P ( as 1 X Q = Q' and P X 0 = P )
PQ + P'Q' ( XOR Expansion, A X B = AB' + A'B )
This is the final simplified expression.
Now we need to check for the options.
If we simplify option D expression.
( P X Q )' = ( PQ' + P'Q )' ( XOR Expansion, A X B = AB' + A'B )
((PQ')'.(P'Q)') ( De Morgan's law )
( P'+ Q).(P + Q') ( De Morgan's law )
P'P + PQ + P'Q' + QQ'
PQ + P'Q' ( as PP' = 0 and QQ' = 0 )
Hence both the equations are same. Therefore Option D.