Two thin lenses when in contact produce a combination of power $+10$ diopter. When they are $0.25 \mathrm{m}$ apart, the power is reduced to $+6$ diopter. The respective power of the lenses in diopter are 

When in contact we have 

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow 10=\frac{1}{f_1}+\frac{1}{f_2}$            $\left\{\because P=\frac{1}{F}=10\right\}$

$\Rightarrow f_1+f_2=10f_1{f}_2$  …….$\left(1\right)$

When at a separation of $0.25\mathrm{m}$ we have, 

$\frac{1}{F^{\text{'}}}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{x}{f_1{f}_2}$

$\Rightarrow 6=\frac{1}{f_1}+\frac{1}{f_2}-\frac{0.25}{f_1{f}_2}$

$\Rightarrow 6=10-\frac{0.25}{f_1{f}_2}$

$\Rightarrow f_1{f}_2=\frac{1}{16}$  ….$.\left(2\right)$

From $\left(1\right)$ we get, 

$f_1+f_2=\frac{10}{16}=\frac{5}{8}$

Now, from $\left(2\right)$ we get, 

$f_1\left(\frac{5}{8}-f_1\right)=\frac{1}{16}$

$\Rightarrow 16f_1^2-10f_1+1=0$

$\Rightarrow 16f_1^2-8f_1-2f_1+1=0$

$\Rightarrow 8f_1\left(2f_1-1\right)-1\left(2f_1-1\right)=0$

$\Rightarrow \left(8f_1-1\right)\left(2f_1-1\right)=0$

$\Rightarrow f_1=\frac{1}{8}\text{ or }{f}_1=\frac{1}{2}$

$\Rightarrow P_1=8D\text{ or }{P}_1=2D$

Hence the correct answer is $\left(2\right)$.