Reduction of chlorate with phosphorus
The number of moles produced for each product will be the number of moles times of the limiting reagent times the reaction stoichiometry (the coefficients).

So to solve this you determine the number of moles of $\ce{KClO3}$ and use that as the base for determining the amount of $\ce{P4O10}$ and the moles of $\ce{KCl}$ produced.  Then multiple by the molecular weight per usual to get your answer.