The oscillation of the plank and block can be considered in 2 parts.

In the SHM part of the motion the block is in contact with one of the 2 springs. The springs are never extended, only compressed, but as one spring is compressed the forces between the block and plank are the same as if the other spring were extended. So ignoring what happens in the gap, the system is equivalent to a two mass one spring system. The angular frequency of this motion is given by $\omega^2=\frac{k}{\mu}$ where $\mu=\frac{mM}{m+M}$ is the reduced mass. The period of this oscillation is $T=\frac{2\pi}{\omega}$.

In the other part of the motion the block and plank are moving towards each other with constant velocity, starting from a distance $3\textit{l}$ apart, because that is the separation of the relaxed springs. (I assume the block has negligible width, otherwise this separation must be reduced by the width of the block.) There is no PE stored in the spring at this stage, so the total KE now equals the initial PE :

$\frac12k\textit{l}^2=\frac12mv^2+\frac12MV^2$.

Momentum is conserved so $MV=mv$ hence $MV^2=(\frac{m}{M})mv^2$. The relative velocity of approach between block and the far spring is

$v+V=v(1+\frac{m}{M})=mv/\mu$.

Therefore

$k\textit{l}^2=(1+\frac{m}{M})mv^2=m^2v^2/\mu$

$(mv)^2=\mu k \textit{l}^2=\frac{k}{\mu}(\mu\textit{l})^2$

$mv=\omega \mu\textit{l}$.

The time taken to cover the round-trip distance of $2\times 3\textit{l}$ between the springs is

$t=\frac{6\textit{l}}{v+V}=\frac{6\textit{l} \mu}{mv}=\frac{6\textit{l} \mu}{\omega \mu \textit{l}}=\frac{6}{\omega}$.

Therefore the total period of the oscillation is $T+t=(2\pi+6)\frac{1}{\omega}$.