# The structure of $N(CH_3)_3$ and $N(SiH_3)_3$ are respectively

$\begin{array}{1 1}(a)\;\text{trigonal planar and pyramidal}\\(b)\;\text{ pyramidal and trigonal planar}\\(c)\;\text{ both pyramidal}\\(d)\;\text{ both trigonal planar}\end{array}$

pyramidal and trigonal planar
$(CH_3)_3N$ has $sp_3$ hybridization on nitrogen with pyramidal shape due to $p\pi-p\pi$ bonding...whereas in $(SiH_3)_3N$ Si has vacant d orbitals which provides $p\pi-d\pi$ bonding and so nitrogen has $sp^2$ hybridization with trigonal planar shape..
Hence (b) is the correct answer.
edited Nov 14, 2017