Explanation

(b)
In $K_{3}\left\lbrack Co\left( CO_{3} \right)_{3} \right\rbrack,\$cobalt shows the +3 oxidation state i.e., $(d^{6})$ ion. Hence, Co (+3) has four unpaired electrons so, it is paramagnetic.
The magnetic moment of Co(+3)
In $K_{3}\left\lbrack Co\left( CO_{3} \right)_{3} \right\rbrack = \sqrt{n(n + 2)}\ BM$
$= \sqrt{4(4 + 2)}$ BM=4.9 BM
Where, n=number of unpaired electrons
$CO_{3}^{2 -}$ is a weak field bidentate ligand, so $3CO_{3}^{2 -}$ ligands occupy six orbitals, thus it shows $sp^{3}\ d^{2}$ hybridisation and octahedral in shape.