When ${\mathrm{PH}}_3$ absorbed in ${\mathrm{HgCl}}_2$ solution the corresponding phosphide is obtained :

When phosphine is exposed to light, it breaks down in water to produce red phosphorous and ${\mathrm{H}}_2$, and when it is absorbed into a solution of copper sulphate or mercuric chloride, the matching phosphides are produced. 

$3{\mathrm{CuSO}}_4+2{\mathrm{PH}}_3\to {\mathrm{Cu}}_3{\mathrm{P}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4$

${\mathrm{Hg}}_3{\mathrm{P}}_2$ is created when ${\mathrm{PH}}_3$ is absorbed by the ${\mathrm{HgCl}}_2$ solution.

$3{\mathrm{HgCl}}_2+2{\mathrm{PH}}_3\to {\mathrm{Hg}}_3{\mathrm{P}}_2+6\mathrm{HCl}$

It produces precipitate of mercury phosphide 

Hence, option B is correct.