0 Dipole Moment!

We need to look at the structure of the substituents in order to find the dipole moment of each molecule. In cases A, B, C, the substituents of the benzene ring aren't linear, so we can have two geometric isomers in each case. For A, the two isomers are:

The right isomer has zero dipole moment since all dipole moments cancel out. However, the left isomer has some non-zero dipole moment, because the component of the dipole moment in -OH add up. Thus, it is possible for A, B, and C to have non-zero dipole moment

However, in D, the carbon atoms in the -CN group are sp hybridized, and are therefore linear.

Since D is symmetric, the dipole moments cancel out, and the net dipole moment of the molecule is zero.

Out of the given four options only $1,4 - dicyanobenzene$ doesn't exhibit $trans$ and $cis$ isomers. Therefore, it is the only aromatic compound that will always have $\mu = 0$ . The thing to note right here is that the substituents are the same and one of the substituent is in $para$ position w.r.t. the other substituent. $Ortho$ and $meta$ positions will never exhibit $\mu = 0$ (I highlighted the word 'always' because you have to mention that in your problem because the $trans$ isomers of the other three aromatic compounds will have $\mu = 0$ ).

Only Carbon doesn't has any lone pair.So dipole moment is 0 in D option.