So after thinking about it a bit, and from the Wikipedia info, the following are my thoughts. If anyone has anything further to add, please do so.
As the electron geometry is tetrahedral (the molecular geometry is trigonal pyramidal), you might expect the P bonding orbitals to be sp3 hybridized, in accordance to the bond angles.
However, note that because of the following (from Wikipedia) :
>>> PH3 is a trigonal pyramidal molecule with C3v molecular symmetry. The length of the P-H bond 1.42 Å, the H-P-H bond angles are 93.5°. The dipole moment is 0.58 D, which increases with substitution of methyl groups in the series: CH3PH2, 1.10 D; (CH3)2PH, 1.23 D; (CH3)3P, 1.19 D. In contrast, the dipole moments of amines decrease with substitution, starting with ammonia, which has a dipole moment of 1.47 D. The low dipole moment and almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals. The high positive chemical shift of the P atom in31P NMR spectrum accords with the conclusion that the lone pair electrons occupy the 3s orbital and so are close to the P atom. This electronic structure leads to a lack of nucleophilicity and an inability to form hydrogen bonds. <<<
(edited : For those who might be confused, I will give a simplified explanation of the statement "the almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals". Recall that p orbitals are px, py and pz, 90 deg angles. Whilst sp3 hybridized orbitals have 109.5 deg angles. Since in PH3 the angles are much closer to 90 deg than 109.5 deg, you would expect the P-H bonds to be mainly sigma bonds from the overlap of p electrons (from P) and s electrons (from H). Notice then, that the s orbitals are largely not involved in bonding, ie. the lone pair occupies the s orbital. Recall that the electrons in the s orbital, on average, are closer (ie. spend more time nearer) to the nucleus, compared to the electrons in the p orbitals. In conclusion, the lone pair is close to the P nucleus, and hence P has slightly increased electron density, which has an effect on polarity of the molecule.)
Hence, due to the particular nature of the PH3 molecular orbitals and the trigonal pyramidal molecular geometry, specifically that the lone pair occupying the 3s orbital and thus being relatively closer to the P atom than the bond pairs between P and H, there is consequently a slightly greater electron density and a slightly partial negative charge on the P atom (even as P and H have the same electronegativity), and hence a slight dipole moment, making PH3 a slightly polar molecule.
Therefore, you would expect the permanent dipole-dipole interactions to be weak, unlike the stronger hydrogen bonding in NH3. Induced dipole-dipole van der Waals interactions, that are present between all molecules (whether polar or non-polar), would play a significant role in the intermolecular attractive forces for PH3.
If anyone has anything further to add, please do so. Thanks in advance!