Let's analyze the geometry of each molecule:

1. $\mathrm{N}({\mathrm{SiH}}_3{)}_3$

In $\mathrm{N}({\mathrm{SiH}}_3{)}_3$, the nitrogen atom is bonded to three ${\mathrm{SiH}}_3$ groups. Nitrogen has a lone pair of electrons. However, due to the presence of vacant d-orbitals in silicon, there is a possibility of pπ-dπ backbonding between the lone pair on nitrogen and the vacant d-orbitals of silicon. This backbonding delocalizes the lone pair, making the nitrogen atom sp² hybridized and resulting in a planar geometry.

2. ${\mathrm{Me}}_3\mathrm{N}$ (Trimethylamine)

In ${\mathrm{Me}}_3\mathrm{N}$, the nitrogen atom is bonded to three methyl groups and has one lone pair of electrons. There is no possibility of backbonding as carbon does not have vacant d-orbitals. Therefore, the nitrogen atom is sp³ hybridized, and the molecule adopts a pyramidal geometry due to the repulsion between the lone pair and bond pairs.

3. $({\mathrm{SiH}}_3{)}_3\mathrm{P}$

In $({\mathrm{SiH}}_3{)}_3\mathrm{P}$, the phosphorus atom is bonded to three ${\mathrm{SiH}}_3$ groups and has one lone pair of electrons. Similar to nitrogen in $\mathrm{N}({\mathrm{SiH}}_3{)}_3$, phosphorus also has vacant d-orbitals, and silicon also has vacant d-orbitals. However, the tendency for pπ-dπ backbonding is much weaker for phosphorus compared to nitrogen because phosphorus is a larger atom, and its lone pair is in a 3p orbital, which is less effective in overlapping with the 3d orbitals of silicon. Therefore, the lone pair on phosphorus remains localized, leading to sp³ hybridization and a pyramidal geometry.

Option Analysis:

• A) planar, pyramidal, planar: Incorrect, as $({\mathrm{SiH}}_3{)}_3\mathrm{P}$ is pyramidal.
• B) planar, pyramidal, pyramidal: Correct, as determined above.
• C) pyramidal, pyramidal, pyramidal: Incorrect, as $\mathrm{N}({\mathrm{SiH}}_3{)}_3$ is planar.
• D) pyramidal, planar, pyramidal: Incorrect, as $\mathrm{N}({\mathrm{SiH}}_3{)}_3$ is planar and ${\mathrm{Me}}_3\mathrm{N}$ is pyramidal.

Correct Answer: (B)