The trouble is because you assumed that the final velocity of the small block is $\sqrt{2gh}$. This is true only if the wedge was stationary (in a frame of reference that is inertial), then what happens is that that the normal force from the wedge on the mass completely balances $mg\cos\alpha$, leaving the component $mg\sin\alpha$ down the wedge as you said.
But the situation is a little more complicated now, because the wedge is moving simultaneously as the small block slides down. So the forces don't balance out as described in the previous paragraph.
One can look at it in terms of energy to gain a better idea. The earth-wedge-mass system is isolated, so its total energy is conserved. The wedge doesn't gain or lose any potential energy, so the only change in potential energy comes from the mass. The change is $- mgh$. This must be distributed to the kinetic energies of BOTH the wedge and the mass. That is,
\begin{align}
&\Delta K + \Delta U = \Delta E = 0 \nonumber \\
\implies & \Delta K_{wedge} + \Delta K_{block} - mgh = 0.
\end{align}
if the wedge wasn't moving at all, we would then have $\Delta K_{wedge} = 0$, so
\begin{align}
\frac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh}
\end{align}
like you said. But we see that if the wedge was moving, it 'eats' up some of the potential energy that would otherwise have gone to the mass. In other words, the small mass' speed will NOT be $v = \sqrt{2gh}$ at the bottom.
Having identified the flaw in your argument, how do we solve the question? There are a few ways. You can draw your force diagrams, carefully balancing out the forces and finding the geometric relation how the position of the mass relates to the position of the wedge. This analysis is perhaps easier in the wedge's frame of reference, but then you would have to add a fictitious force as it is not an inertial frame.
But the easiest analysis would be in terms of energy conservation, like the equation I gave you. We have
\begin{align}
\frac{1}{2}MV^2 + \frac{1}{2}mv^2 - mgh = 0.
\end{align}
Now all you have to do is find how $V$ is related to $v$. This is simple from conservation of momentum and some trigonometry, try it.
(Edit)
I noticed after posting that you specifically highlighted the fact that $v = \sqrt{2gh}$ is with respect to the wedge. Lest you start pointing that out, this is not true, because the force the mass feels down the wedge is not $mg\sin\alpha$, because in this frame (wedge's frame, which is not inertial), there is the fictitious force.
In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here.
Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not the true static friction force. In this case, there is no other acceleration, so there is no need for static friction. Static friction only comes into play when the two bodies are attempting to be in relative motion with each other. This is not the case here, at the point of contact the velocities of the corresponding points on the wheel and platform are equal and there is no force trying to stop this.
When you're standing on the ground, you're not mysteriously being pushed by friction. It's the same thing here, the wheel is "standing" with respect to the point of contact, though the points of contact are changing over time.
Best Answer
The block's $x$-velocity right before it hits the floor is not $v_b$, but $v_b \cos \alpha + v_I$. The $\cos \alpha$ is because it's not moving horizontally relative to the wedge, but at an angle $\alpha$; the $+ v_I$ term is because it's moving relative to the wedge, and momentum is only conserved in a frame that is at rest with respect to the table. Similarly, its $y$-velocity is $v_b \sin \alpha$. So the correct equations are (taking into account the relative motions): $$ \frac{1}{2} m (v_b^2 + v_I^2 + 2 v_b v_I \cos \alpha) + \frac{1}{2} M v_I^2 = mgh $$ $$ m (v_b \cos \alpha + v_I) + M v_I = 0 $$ Solving these two equations yields the expression you got from applying Newton's Second Law.