In solution of methanol $\&$ benzene methanol molecules are held together due to hydrogen bonding as shown below.

$\begin{array}{*{20}{c}} {\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}\\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,}\\ { - - - - OH - - - - OH - - - - OH - - - - } \end{array}$

On adding benzene, the benzene molecules get in between the molecule of methanol thus breaking the hydrogen bonds. As the resulting solution has weak intermolecular attraction, the escaping tendency of alcohol $\&$ benzene molecule from the solution increases. Consequently the vapour pressure of the solution is greater than the vapour pressure as expected from Raoult's law.