$Ca_{3}(PO_{4})_{2}$

Calcium phosphate

$Ca_{3}(PO_{4})_{2}$

So,Its Molecular mass

$=40\times 3+(31+16\times 4)\times 2=310$

Molecular mass of $P_{2}O_{5}$

$=(31\times 2+16\times 5)=142$

Now,

$Ca_{3}(PO_{4})_{2}\longrightarrow 3CaO+P_{2}O_{5}$

Now In 1 molecule of Calcium phosphate there is only one molecule of Phosphorus pentoxide( $P_{2}O_{5}$ )

in $310$ molecular mass of Calcium phosphate the amount of $P_{2}O_{5}$ is $142$

So, in 100 molecular mass of Calcium phosphate the amount of $P_{2}O_{5}$ is

$\dfrac{142\times 100}{310}=45.8064516129 \%$