The correct statement among the following is:

We start by recalling the general shape of an amine. For a simple trialkylamine like $$\mathrm{(CH_3)_3N}$$ the nitrogen is $$\mathrm{sp^3}$$-hybridised. Three of the hybrid orbitals contain $$\sigma$$-bonding pairs (one to each carbon) and the fourth contains the lone pair. With four regions of electron density, VSEPR predicts a tetrahedral arrangement of orbitals and, because one of those regions is a lone pair, the observed molecular geometry becomes pyramidal. Thus $$\mathrm{(CH_3)_3N}$$ is pyramidal.

Now we examine $$\mathrm{(SiH_3)_3N}$$. Each silicon atom possesses energetically accessible empty 3d orbitals, while the nitrogen atom has a filled $$\mathrm{p}$$-type lone pair. A lone pair on nitrogen can overlap sideways with the vacant 3d orbitals on the three silicon atoms. This interaction is described as $$p\!\!-\!d$$ $$\pi$$ back-bonding. Whenever such delocalisation occurs, the requirement for a localised tetrahedral lone pair on nitrogen is removed, and the three $$\sigma$$ bonds together with the delocalised lone pair adopt a trigonal arrangement. In other words, the molecule flattens out at nitrogen to give a trigonal-planar shape. Therefore $$\mathrm{(SiH_3)_3N}$$ is planar.

Basicity depends on the availability of the lone pair for protonation. In $$\mathrm{(CH_3)_3N}$$ the lone pair is localised and free to donate; its electron density is concentrated on nitrogen, so the molecule is reasonably basic. In $$\mathrm{(SiH_3)_3N}$$ the lone pair is partially withdrawn from nitrogen into the $$p\!\!-\!d$$ $$\pi$$ system. Delocalisation lowers the electron density on nitrogen and makes the lone pair less available for bonding to $$\mathrm{H^+}$$. Consequently its basic strength decreases.

Putting the two pieces together, we have:

$$\mathrm{(SiH_3)_3N}$$ is planar  and  less basic than $$\mathrm{(CH_3)_3N}$$.

Option A states exactly this relationship.

Hence, the correct answer is Option A.