b
When resistances $4\,\Omega $ and $12\,\Omega $ are connected in series $ = 4 + 12 = 16\,\Omega $
When these resistance are connected in parallel.
$\frac{1}{{{R_P}}} = \frac{1}{4} + \frac{1}{{12}}$ $ \Rightarrow \,\,\,{R_P} = \frac{{4 \times 12}}{{4 + 12}} = \frac{{4 \times 12}}{{16}} = 3\,\Omega $