Why does monoammonium phosphate (MAP) have 52% of phosphorus (P₂O₅)?
When it comes to fertilizers, usually what is advertised by general name monoammonium phosphate 10-52-0, in reality is ammophos, which is a mix of monoammonium and diammonium phosphates, $\ce{NH4H2PO4}$ and $\ce{(NH4)HPO4}$ with some water, so the brutto formula is 

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}$$

where $0 ≤ x ≤ 1$.
To find out the range for $y$, let's find the average molecular mass from the given nitrogen ($\ce{N}$) and phosphorous ($\ce{P2O5}$) content $α_i$:

$$\bar{M} = \frac{1}{2}\cdot\left(\frac{M(\ce{N})}{α(\ce{N})} + \frac{M(\ce{P2O5})}{2α(\ce{P2O5})}\right) = \frac{1}{2}\cdot\left(\frac{\pu{14 g mol-1}}{0.10} + \frac{\pu{142 g mol-1}}{2\cdot 0.52}\right) \approx \pu{138 g mol-1}$$

Using the expression for molecular mass of brutto formula and its average value:

$$x = 1.35 - 1.06y$$

Since $0 ≤ x ≤ 1$, the following holds true for $y$:

$$0.33 ≤ y ≤ 1.27$$

resulting in formula with the following ranges for the variables:

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}, \quad x \in [0; 1], y\in [0.33; 1.27]$$

Since monoammonium phosphate usually prevails in this mix, there is a substantial amount of water; e.g. one of the possible brutto-formulas would be the following monohydrate:

$$\ce{(NH4)_{1.3}H_{1.7}PO4 * H2O}$$

Obviously, all the above is valid in the absence of third-party contaminations, e.g. if pure phosphoric acid and ammonia were used for the synthesis.