$1$ mole of each of $Ca(OH{)}_2$ and $H_{3}P{O}_4$ are allowed to react under dilute conditions. The maximum number of moles of $C{a}_3(P{O}_4{)}_2$ formed is 

 $3Ca(OH{)}_2+2H_{3}P{O}_4\to C{a}_3(P{O}_4{)}_2+6H_{2}O$

Here, we can see that $3$ moles of $Ca(OH{)}_2$ is reacting with $2$moles of $H_{3}P{O}_4$.

But it is given that $1$mole of both the reactants is reacting.

If $1$mole of $Ca(OH{)}_2$ is taken then, $0.667$ moles of $H_{3}P{O}_4$ is required.

But it is given that $1$mole of  $H_{3}P{O}_4$ is reacting.

Hence, $H_{3}P{O}_4$ is present in excess.

Now, from the reaction 3 moles of $Ca(OH{)}_2$  forms $1$mole of $C{a}_3(P{O}_4{)}_2$

$1$mole of $Ca(OH{)}_2$  will produce $1/3$ moles of $C{a}_3(P{O}_4{)}_2$

Therefore, Option (C) is correct.