(b) ${\overrightarrow{E}}_{1} = \frac{1}{4\mu\pi\varepsilon_{0}}.\frac{2\overrightarrow{P}}{r^{3}}$ and
${\overrightarrow{E}}_{2} = - \frac{1}{4\pi\varepsilon_{0}}.\ \frac{\overrightarrow{P}}{{(2r}^{3})} = - \frac{1}{4\pi\varepsilon_{0}}.\frac{\overrightarrow{P}}{{8r}^{3}}$
$\Rightarrow \ \ {\overrightarrow{E}}_{2} = - \frac{{\overrightarrow{E}}_{1}}{16}$
(Here negative sign means direction)