Vapour pressure and Raoult's Law
During the process of distillation, we condense the vapours (B) of the mixture (A) to give a mixture (C) that has the same mole fractions as that of the vapour (B) of mixture (A).
In order to solve this question there would be three steps.

Firstly, finding the vapour pressure of the initial mixture.
Second, we find the mole fraction of components in the vapour phase of this mixture.
The third and final part would be to find the vapour pressure of the distillate leading to our answer.

However, finding the actual composition of the vapours here, would be the biggest challenge in this question.
Step $1$
In order to find the vapour pressure, we use Raoult's Law which mathematically states:
$$P_T = \sum{P^\circ_i\chi_i}$$
Since this is a two-component system, the total pressure would be expressed as a sum of two terms for each component A and B.
$$P_T = P^\circ_\mathrm a \chi_\mathrm a +P^\circ_\mathrm b \chi_b$$
Substituting the values given in the question, we get $P_T= \pu{85 mmHg}$
Step $2$
The next step is to find the composition in the vapour phase. For this, we use the  property that the partial pressure of a component in the vapour phase is equal to the contribution of the same component in the liquid phase. Mathematically, it is equivalent to stating:
$$P_Ty_\mathrm a= P^\circ_\mathrm a \chi_a$$
Here, $y_a$ is the mole fraction of component A in vapour phase. Solving for $y_\mathrm a$, we get that $y_\mathrm a = \frac{5}{17}$, which implies $y_\mathrm b = \frac{12}{17}$.
Step $3$
The final is exactly the same as the first step, wherein you use Raoult's law to find the final vapour pressure of the distillate (mole fraction of components equal to that mole fraction of vapour in initial mixture).
Solving for the vapour pressure, you get $P_T^{'} = \pu{85.88 mmHg}$, which is the answer provided.