The correct relationship between molarity ($\mathrm{M}$) and molality $\left(\mathrm{m}\right)$ is $(\mathrm{d}=$ density of the solution, in ${\mathrm{KgL}}^{-1},{\mathrm{M}}_2=$ molar mass of the solute in $\mathrm{kg}$ ${\mathrm{mol}}^{-1}$ )

We know that

Molarity (M)$=\frac{No. of moles of solute}{Volume of solution\{ in litres\}}$

Molality (m) $=\frac{No. of moles of solute}{weight of solvent}$

If a be density in $\mathrm{g}/{\mathrm{cm}}^3$.

Then mass of the solution $=1000\mathrm{d} \mathrm{gm}$

mass of solute $=M\times M_2$

where $M_2=$ molar mass of solute

Mass of solvent $=1000 \mathrm{d}-{\mathrm{MXM}}_2$

$\therefore Molality\left(m\right)=\frac{M\times 1000}{1000d-M\times M_2}$

$\therefore M=\frac{md}{1+m{M}_2}$

Hence, option (a) is the answer.