In the view of electronegativity, P-O-H loses its proton easily whereas in the point of bond length, P-H will lose its proton quickly. So that doubt arises now. Among these two factors, which one decide the fate?
I think that electronegativity must be considered first.
Well, as the radius of the atom increases electronegativity decreases and as the atom's radius increases the bond length increases, what is concluded is the following: as the bond length increases electronegativity decreases.
So for the proton, when the bond length increases, the H+ will be releases easily but the electronegativity decreases so the H+ will not be released easily. So they don't work together in the case of the proton.
Because as we know, if the electron of the H is more attracted to the atom which the H is bonded to, then the H+ will be released easily (compare HF and HCl) which makes sense... so electronegativity is stronger than the factor of bond length.
I think the electronegativity in this case has a larger component of responsibility in the loss of the proton ... a point for this is the amount of available oxygen that electrons can coordinate the hydrogen and / or metal ... an analysis of the atoms electronegativities present in these groups can clarify this issue.
In POH the H is a proton whereas in P-H it is a hydride. There is a significant difference. Are you simply referring to the loss of H or are you specific about it being lost as H+ ?
According to theoretical analysis carried out (attachment) for systems PH3 and PH2OH, the energy needs for dissociation of the (P)O-H bond is lower than, this one for the dissociation of the (H2)P-H bond.
Bellow you can find additional computational data.