Given, refractive index of lens, $\mu_g=1.5$
Refractive index of medium, $\mu_m=1.6$
Power of lens, $P=-5 \mathrm{D}$
Focal length of lens, $f_l=\frac{1}{-5} \mathrm{~m}=-20 \mathrm{~cm}$
From lens maker formula, we have
$$
\begin{array}{r}
P=\frac{1}{f_l}=\left(\frac{\mu_s}{\mu_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \quad \ldots \text { (i) } \\
\left(\because \mu_a=\right.\text { refractive index of air) }
\end{array}
$$
$$
P^{\prime}=\frac{1}{f_m}=\left(\frac{\mu_g}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(ii)
$$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{aligned}
\frac{P^{\prime}}{P} & =\frac{\left(\frac{\mu_g}{\mu_m}-1\right)}{\left(\frac{\mu_g}{\mu_a}-1\right)}=\frac{\frac{1.5}{1.6}-1}{\frac{1.5}{1}-1} \\
\Rightarrow \quad & \frac{P^{\prime}}{P}=\frac{-1}{16} \times \frac{2}{1} \Rightarrow P^{\prime}=\frac{-1}{8} \times(-5) \\
\Rightarrow & P^{\prime}=0.625 \mathrm{D}
\end{aligned}
$$
This is close to option (a).