In ${\mathrm{PO}}_4^{3-}\text{ ion, }$the formal charge on each oxygen atom and $P-O$ bond order respectively are

Bond order $=\frac{\text{ Total number of bonds between atoms }}{\text{ Total number of resonating structure }}$
 

Bond order =5​/4=1.25

In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each.
In the resonance hybrid, a total of -3 charge is distributed over four O atoms. 

Thus the formal charge of each O atom = −3​/4 = −0.75.