When particle is displaced by a distance x in the downward direction, lower spring is compressed by $\text{x}$ distance in downward direction and both upper string is displaced by $\text{xcos}\left(45°\right)$ at an angle of $45°$ from vertical.Due to these compressions and expansions, net force start working on the particle which tries to bring the particle back to its original position.

Writing net spring force on the particle-
${\mathrm{F}}_{\mathrm{net}}=\mathrm{F}+{\mathrm{F}}_1\cos \left(45°\right)+{\mathrm{F}}_2\cos \left(45°\right)$
$\Rightarrow {\mathrm{F}}_{\mathrm{net}}=\mathrm{kx}+\left[\mathrm{kxcos}\left(45°\right)\right]\cos \left(45°\right)+\left[\mathrm{kx}\left(\cos 45°\right)\right]\cos \left(45°\right)$
$\Rightarrow {\mathrm{F}}_{\mathrm{net}}=2\mathrm{kx}$

From newton's second law-
$\mathrm{ma}=-2\mathrm{kx}$
$\Rightarrow \mathrm{a}=-\left(\frac{2\mathrm{k}}{\mathrm{m}}\right)\mathrm{x}$

Above equation represents SHM.Comparing it with equation of SHM
$\mathrm{a}=-{\mathrm{\omega }}^2\mathrm{x}$
$\Rightarrow \mathrm{\omega }=\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}$
Since time period $\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}$
 $\therefore \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}}$