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Let $\oplus$ denote the exclusive OR (XOR) operation. Let '1' and '0' denote the binary constants. Consider the following Boolean expression for $F$ over two variables $P$ and $Q$:

$$F(P,Q)=\left( \left(1 \oplus P \right) \oplus \left( P \oplus Q \right )\right ) \oplus \left(\left(P \oplus Q\right) \oplus \left(Q \oplus 0\right)\right)$$

The equivalent expression for $F$ is

(A) $P+Q$

(B) $\overline{P+Q}$

(C) $P \oplus Q$

(D) $\overline {P \oplus Q}$

XOR is associative and commutative. Also, $A \oplus A = 0$ and $A \oplus 1 = \overline{ A}$ and $A \oplus 0 = A$.  So
$\left( \left(1 \oplus P \right) \oplus \left( P \oplus Q \right )\right ) \oplus \left(\left(P \oplus Q\right) \oplus \left(Q \oplus 0\right)\right)$
$\implies \left(1 \oplus P \right) \oplus \left( \left( P \oplus Q \right ) \oplus \left(P \oplus Q \right) \right) \oplus \left(Q \oplus 0\right)$
$\implies \left(1 \oplus 0 \right) \oplus \left( P \oplus Q \right)$
$\implies 1 \oplus \left( P\oplus Q \right)$
$\implies \overline {\left( P \oplus Q\right)}$

edited

D)
Since there are only 2 variables putting in pair of values of P and Q in F and checking with the options is a time saving method.
But Lets solve it.

observe the common term p ex or q in both

consider a case where p and q are equal

then p ex or q results in 0

1)in first p is ex ored  with 1

2)in second q is ex ored with 0

so if p, q are same then either of one oresults in 1 and another to 0

1 ex or  0

it is exnor