The formula for copper(II) phosphate is Cu3(PO4)2. This means that for every 3 copper atoms, there are 2 phosphate ions. Each phosphate ion contains 1 phosphorus atom.

So, to find the number of atoms of phosphorus in 4.40 mol copper(II) phosphate, we can use the following calculation:

$$4.40 \text{ mol Cu}_3(PO_4)_2 \times \frac{2 \text{ mol P}}{1 \text{ mol Cu}_3(PO_4)_2} \times \frac{1 \text{ atom P}}{1 \text{ mol P}} = \boxed{8.80 \times 10^{23} \text{ atoms P}}$$

Therefore, there are 8.80 x 10^23 atoms of phosphorus in 4.40 mol copper(II) phosphate.