Both molecules can be considered the acid (hydrate) trifluoride of the respective acid (phosphoric acid or nitric acid). In the case of phosphoric acid, this is immediately obvious: Just replace the three hydroxide groups with a fluorine atom, each. For the nitrogen compound, you initially need to (formally) add water before you can do the substitution:
$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$
(Note: this is not a proper chemical reaction, merely an illustration of the thought processes.)
You could also approach this compound from the ammonium cation $\ce{NH4+}$ by replacing three hydrogen atoms with fluorine and the fourth with oxygen. Or you start off with $\ce{NF3}$ (isoelectronic to $\ce{PF3}$ and then oxidise the nitrogen by adding a single oxygen atom much in the same way you would oxidise from $\ce{HNO2}$ to $\ce{HNO3}$.
In fact, for many of the compounds traditionally explained with d-orbital participation for main-group elements, an explanation adhering to the octet rule (no more than eight valence electrons per main-group atom; two for hydrogen/helium) is often closer to the truth. This includes, but is not limited to $\ce{OPF3, SO4^2-}$, sulphonic acids and many more.