Lift the second Block!

Since there is e=0, B does not contribute to lifting it up. Only spring energy has to lift it up. After the spring is compressed, A moves up with velocity till its up motion is stopped. So question of lifting A up does not arise, if before that B is lifted up as in our case. So if spring is extended by x with reference to up moving A, so that the spring forces B up, with the force of B's weight, B is just lifted. That implies 2mg=xk. That is $\color{#3D99F6}{x=\dfrac {2mg}{k}}$ . That means, A is( h - x) high at that point. Implies spring must have stored P.E. energy lost by A. But P.E. lost = mass*(gravity acc.) *( height lost ) =Energy gained by the spring due to A. Note:-During this proses, one end of the spring is as if fixed to the ground.
$1*m*g*(h-x)=\dfrac 1 2 *k*x^2 .\\ \implies mg(h-\dfrac {2mg}{k})=\dfrac 1 2 *k*(\dfrac {2mg}{k} )^2. \\Solving~~h=\dfrac{4mg}{k}= \dfrac{nmg}{k}, \implies~n=~~~~~\Large \color{#D61F06}{4}$

$|x_i|=$ maximum compression of spring ... $x_i<0$

$|x_f|=$ maximum elongation of spring ... $x_f>0$

In order for mass B to leave the ground ... $kx_f=2mg$ ...This is the first constraint.

Relationship between $x_f$ and $x_i$ from conservation of energy (all of the energy of block B is lost in the inelastic collision) ... $0.5kx_i^2=0.5kx_f^2+mg(x_f-x_i)$ ... This is the second constraint.

Relationship between $x_i$ and $h$ from conservation of energy ... $0.5kx_i^2=mg(h-x_i)$ ... This is the third and final constraint.

These three constraints solve the problem. To finish, combine the first two constraints to eliminate $x_f$ and you will get $0.5kx_i^2=\frac{4(mg)^2}{k}-mgx_i$ and then set this equal to the third constraint and you will get $h=\frac{4mg}{k}$ which is the minimum value of h.

$kx = mg$

$\Rightarrow x = \frac{2mg}{k}$

Initial energy of A (only) $E_0 = mgh$

Energy at final position (spring stretched a distance $x$ $E_{f-minimum} = \frac{kx^2}{2} + mgx = 2\cdot\frac{2m^2g^2}{k}$

$mgh = \frac{4m^2g^2}{k}$

$\Rightarrow\boxed{~ \boxed{h = 4\cdot \frac{mg}{k}}~}$

@Nishant Rai do u have any shorter method