What would the phosphorus-NMR of trifluorodiphenylphosphorane look like?
Well, these are the fundamentals of NMR that are important:


Signals will be split if they couple to neighbouring atoms with a spin $\ne 0$ (though coupling to spin ½ nuclei is most common especially in organic NMR).
Every different environment of a coupling nucleus will produce a distinct coupling.
Depending on the number of nuclei in a given environment, the number of signals (coupling pattern) is given by the equation $2nI+ 1$, where $n$ is the number of nuclei and $I$ is that nucleus’ spin.


In your example you have correctly identified two different fluorine environments with one and two fluorines in them respectively.

The single equatorial fluorine will give a doublet because $2nI +1 = 2\ (n = 1, I = \frac{1}{2})$.

For the same reason, the axial fluorines will give a triplet: $2nI + 1 = 3\ (n =2, I = \frac{1}{2})$.

If one coupling gives a triplet and the other a doublet, the final result is either a doublet of triplets or a triplet of doublets. (The larger coupling constant i.e. the greater splitting is given first.)



The fluorine NMR would be more complex: They can couple to both phosphorus and the other fluorine. It is important to know that the axial fluorines are homotopic, i.e. the molecule can be rotated exactly exchanging their places — this means that the two give an identical signal. The equatorial one cannot be brought to overlap with the axial fluorines by any mean of symmetry; the two environments are heterotopic, i.e. they will produce different signals (and typically couple).


The axial fluorines will couple to the equatorial fluorine and phosphorus.
The equatorial fluorine will couple to the axial ones and phosphorus.


I’ll leave you to work out the coupling patterns. (Hint: The less bonds between two coupling nuclei, the larger the coupling constant usually is.)



A final note: It is due to electronic reasons that the phenyl groups occupy the equatorial positions rather than going axial. A four-electron three-centre bond is more stable for electronegative elements than electropositive ones.