Two blocks, of masses $M$ and $2M$, are connected to a light spring of spring constanat $K$ that has one end fixed, as shown in figure. Mass of the hanging block is 2M .The horizontal surface and the pulley are frictionless. The blocks are released from when the spring is non deformed. The string is light. 

$$\begin{gathered} Conserving energy , 2Mg\times x_{max}=\frac{1}{2}k{\left(x_{max}\right)}^2 \\ \Rightarrow x_{max} =\frac{4\mathrm{Mg}}{\mathrm{K}} \end{gathered}$$

System will have maximum $KE$ when net force on the system becomes zero. Therefore $2\mathrm{Mg}=\mathrm{T}$and $\mathrm{T}=\mathrm{kx} \Rightarrow \mathrm{x}=\frac{2\mathrm{Mg}}{\mathrm{K}}$

Hence $\mathrm{KE}$ will be maximum when $2\mathrm{M}$ mass has gone down by $\frac{2\mathrm{Mg}}{\mathrm{K}}.$

Applying W/E theorem

$K{E}_{max}-0=2\mathrm{Mg}·\frac{2\mathrm{Mg}}{\mathrm{K}}-\frac{1}{2} \mathrm{K}·\frac{4{\mathrm{M}}^2{ \mathrm{g}}^2}{{ \mathrm{K}}^2};K{E}_{max}=\frac{2{\mathrm{M}}^2{ \mathrm{g}}^2}{{ \mathrm{K}}^2}$

Maximum energy of spring

$\frac{1}{2} \mathrm{K}·{\left(\frac{4\mathrm{Mg}}{\mathrm{K}}\right)}^2=\frac{8{\mathrm{M}}^2{ \mathrm{g}}^2}{ \mathrm{K}}$

Therefore Maximum spring energy = $4\times$ maximum $K.E.$

When $\mathrm{K}.E.$ is maximum $\mathrm{x}=\frac{2\mathrm{Mg}}{\mathrm{K}}.$

Spring energy =$\frac{1}{2}·\mathrm{K}·\frac{4{\mathrm{M}}^2{ \mathrm{g}}^2}{ \mathrm{K}}=\frac{2{\mathrm{M}}^2{ \mathrm{g}}^2}{ \mathrm{K}} i.e.\left(4\right)$ is wrong.