How to calculate the number of moles present in a litre of saturated solution?
For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:

$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$ 

Upon dissolution of one unit of calcium phosphate, we get five units of ions total, true, but the solubility product is defined as follows (from the IUPAC Gold Book):

The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.

For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate). 

Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed. 

$\ce{PO_{4}^{3-} + H_2O -&gt;HOPO_3^{2-} + HO^-}$ 

Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion. 

Therefore, based on the IUPAC definition:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$ 

Taking $\ce{x}$ to be one unit of calcium ion:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$ 

Solving the above yields $x = 10^{-6}$. 

Therefore, with respect to your original question, the number of moles of calcium phosphate present initially can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer.