Problem
The correct statement is
(A) P & Q are aromatic compound and Q has $\mathrm{sp^3}$-hybridized carbon atom.
(B) P is aromatic with 10 π electrons and Q is aromatic with 2 π electrons.
(C) P is aromatic and Q is anti-aromatic.
(D) P is aromatic and Q is non-aromatic.
Answer
(A) P & Q are aromatic compound and Q has sp3-hybridized carbon atom.
Question
I got P easily; the two $\ce{K}$ atoms would lose two electrons and form the cyclooctatetraene dianion, which will become planar and aromatic with 10 π electrons.
However, I could not find what Q will be. My first step was a proton getting attached to one carbon of any of the four $\ce{C=C}$, which will give $\mathrm{sp^3}$ hybridization on that carbon which accepted the proton. There will now be another adjacent carbon with a positive charge on it, which will be delocalized through the 6 π bonds.
However, the presence of $\mathrm{sp^3}$ carbon in between will prevent complete conjugation throughout the ring.
I don't think the next step can be the attack of $\ce{HSO4^-}$ as a nucleophile, because $\ce{HSO4^-}$ is a stable anion and hence a very poor nucleophile. Water isn't mentioned, so I couldn't assume attack of $\ce{OH^-}$ as a nucleophile either.
In any case, I don't see how Q can be aromatic. What am I missing?