At this stage you're just working out how many moles of compound you have, irrespective of the reaction stoichiometry. MW = 151.13 g/mol, so
nC7H5NO3 = 0.25 g C7H5NO3 (1 mol C7H5NO3/151.13 g C7H5NO3)
is correct.
The stoichiometry is taken care of by the factor (1 mol/2 mol) in your next equation - i.e. you expect 1 mol indigo for every 2 mol of precursor.