In this situation, if mass m moves downwards a distance $x$ from the equilibrium position, the pulley will also move by x and so, the spring will stretch by $2x$.

Therefore, the spring force will be $2kx$. The restoring force on the block will be $4 kx$.

Hence, $F=- 4 kx$



$\Rightarrow mg-4kx=ma$

$\Rightarrow ma=-4kx+mg$

$\Rightarrow$    $a=-\frac{4kx}{m}$ $+g\Rightarrow {\omega }^2=\left|\frac{a}{x}\right|$

                $T=2\text{π}\sqrt{\left|\frac{x}{a}\right|}$

                $T=2\text{π}\sqrt{\left(\frac{m}{4k}\right)}$