The geometry with respect to the central atom of the following molecules respectively are :

$\mathrm{N}{\left({\mathrm{SiH}}_3\right)}_3;{\mathrm{Me}}_3\mathrm{N};{\left({\mathrm{SiH}}_3\right)}_3\mathrm{P}$

$\mathrm{N}{\left({\mathrm{SiH}}_3\right)}_3$ is planar because N is ${\mathrm{sp}}^2$ hydridized with a trigonal planer shape, leaving only one N in the P orbit and a vacant D orbit for each si. As a result, there is p(pi)-d(pi) block bonding, which stabilises the structure.
${\left({\mathrm{SiH}}_3\right)}_3\mathrm{P}$
Pyramidal is 3P.
The pyramidal ${\mathrm{Me}}_3\mathrm{N}$