Step-by-step Solution:

1. Analyze the Structures and Hybridization of NH₃ and PH₃:

   • Ammonia (NH₃): The central atom is Nitrogen (N). The electronic configuration of N is [He] 2s² 2p³. In NH₃, nitrogen forms three single bonds with hydrogen atoms and has one lone pair of electrons. The steric number is 4 (3 bond pairs + 1 lone pair). According to VSEPR theory, this corresponds to a tetrahedral electron geometry and a trigonal pyramidal molecular shape. The H-N-H bond angle is approximately 107.8°, which is close to the ideal tetrahedral angle of 109.5°. This indicates that the nitrogen atom is approximately sp³ hybridized. The lone pair resides in one of the four sp³ hybrid orbitals.
   • Phosphine (PH₃): The central atom is Phosphorus (P). The electronic configuration of P is [Ne] 3s² 3p³. Similar to NH₃, phosphorus forms three single bonds with hydrogen atoms and has one lone pair. However, the H-P-H bond angle is about 93.5°. This angle is very close to 90°, which is the angle between pure p-orbitals. This suggests that in PH₃, there is very little hybridization (or it can be explained by Drago's rule). The P-H bonds are formed mainly by the overlap of the three 3p orbitals of phosphorus with the 1s orbitals of hydrogen. Consequently, the lone pair of electrons resides in an orbital with a very high s-character, which is nearly the pure 3s orbital.

2. Compare the Electron Donation Capability (Lewis Basicity):

   • A Lewis base is a substance that can donate a pair of electrons. The ease of donation depends on the availability and orientation of the lone pair.
   • In NH₃: The lone pair is in an sp³ hybrid orbital. An sp³ orbital has 25% s-character and 75% p-character. The significant p-character makes the orbital highly directional, extending out into space away from the N-H bonds. This directional nature facilitates effective overlap with an empty orbital of a Lewis acid (like a proton or a metal ion).
   • In PH₃: The lone pair is in an orbital with high s-character (almost pure 3s). An s-orbital is spherical and non-directional. Furthermore, electrons in an s-orbital are held more tightly by the nucleus compared to electrons in p or hybrid orbitals of the same principal quantum number. This makes the lone pair in PH₃ less available for donation and less effective in forming a directional coordinate bond.

3. Conclusion:

   • Due to the directional nature of the sp³ hybrid orbital containing the lone pair, ammonia (NH₃) is a much better electron donor (stronger Lewis base) than phosphine (PH₃).
   • The lone pair in PH₃ is less available and less directional because it resides in an orbital with high s-character.

4. Evaluate the Options:

   • A: Between NH₃ and PH₃, NH₃ is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional. This statement correctly identifies NH₃ as the better donor but gives the wrong reason. The lone pair in NH₃ is in a directional sp³ orbital, not a spherical 's' orbital.
   • B: Between NH₃ and PH₃, PH₃ is a better electron donor because the lone pair of electrons occupies sp³ orbital and is more directional. This statement is incorrect. NH₃ is the better donor, and the lone pair in PH₃ has high s-character, not sp³.
   • C: Between NH₃ and PH₃, NH₃ is a better electron donor because the lone pair of electrons occupies sp³ orbital and is more directional. This statement is correct. It correctly identifies NH₃ as the better donor and provides the correct reason based on the sp³ hybridization and the resulting directionality of the lone pair orbital.
   • D: Between NH₃ and PH₃, PH₃ is better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional. This statement is incorrect. PH₃ is a weaker donor precisely because its lone pair occupies an orbital with high s-character which is less directional.