Why does the PH3 geometry deviate more from the trigonal planar one, than does NH3?
Apparantly it was community consensus that this question be clear enough to stay open.

Neither $\ce{PH3}$ nor $\ce{NH3}$ have any incentive whatsoever to assume a trigonal planar geometry. The most stable ground state of both nitrogen and phosphorus atoms is $n\mathrm{s^2},\ n\mathrm{p^3}$ where $n = 2$ for nitrogen and $n=3$ for phosphorus. Since the p-orbitals are higher in energy per se, it makes sense to use them for bonding and lower their energy — which, as a side note leads to much more efficient overlap than using s-orbitals for bonding, since p-orbitals have a direction in space.

Therefore, the most stable geometry for both would be trigonal pyramidal with $90^\circ$ angles between the substituents. This would allow all-p bonding orbitals and an all-s lone pair on the central atom. If you check out $\ce{PH3}$, you will find almost exactly that geometry save a little distortion.

The distortion you see in $\ce{PH3}$ is of the same type as the one which moves the hydrogen atoms much further apart in $\ce{NH3}$: Steric stress. The hydrogens themselves repel each other; they require a certain space. Around a large phosphorus atom there is enough space for them to crowd at the ideal bonding angle while the small nitrogen does not offer that space. The final bond angle of $\approx 107^\circ$ is a fine-tuned result of trying to keep the hydrogen atoms as far away from each other as necessary and keeping the bonding angle as close to $90^\circ$ as possible.

$\ce{BH3}$ is an odd one out in this comparison. Boron does not have a lone pair and thus we expect isolated $\ce{BH3}$ to assume trigonal planar geometry. However, we also expect isolated $\ce{BH3}$ to have a low stability since boron does not have an octet. Hence it dimerises to $\ce{B2H6}$ with two $\ce{B-H-B}$ banana bonds of the 2-electron-3-centre type. The result can be described as a distorted tetrahedron around boron.