Consider Option(D). LHS can be simplified as,
$$(P \oplus \overline{P}) \oplus Q = 1 \oplus Q = \overline{Q}$$
Similarly, for RHS
$$(P \odot \overline{P}) \odot \overline{Q} = 0 \odot \overline{Q} = Q$$
$LHS \neq RHS$ therefore, Option (D) is the correct answer.
Other options can be simplified as follows.
- $\overline{P \oplus Q} = \overline{P\overline{Q} + \overline{P}Q} = \left(\overline{P\overline{Q}}\right) \left(\overline{\overline{P}Q}\right) = \left(\overline{P}+Q \right) \left( P+\overline{Q}\right) = PQ + \overline{P}\ \overline{Q} = P \odot Q$
- $\overline{P} \oplus Q = \left( \overline {P}\right)\overline{Q} + \overline{\left( \overline {P}\right)}Q = PQ + \overline{P}\ \overline{Q} = P \odot Q$
- $\overline{P} \oplus\overline{Q} = \left( \overline {P}\right) \overline{\left( \overline {Q}\right)} + \overline{\left( \overline {P}\right)}\left( \overline {Q}\right) = \overline{P}Q + P\overline{Q} = P \oplus Q $