1. Given expression

$S=\alpha^2\beta\ln\left(\frac{\mu kR}{J\beta^2}+3\right)$

Since logarithm's argument must be dimensionless,

$\frac{\mu kR}{J\beta^2}$ must be dimensionless.

2. Dimensions of entropy

Using

$dS=\frac{dQ}{T}$

entropy has dimensions of energy per temperature:

$[S]=\frac{[Q]}{[T]}=\frac{ML^2T^{-2}}{\Theta}=ML^2T^{-2}\Theta^{-1}$

Also, gas constant $R$ has the same dimensions as entropy per mole, so since $\mu$ is number of moles,

$[\mu R]=[S]$

Boltzmann constant $k$ also has dimensions of entropy:

$[k]=[S]$

Hence,

$[k]=[\mu R]=[S]$

3. Find dimensions of $\beta$

From dimensionless logarithm argument,

$\left[\frac{\mu kR}{J\beta^2}\right]=1$

So,

$[\beta^2]=\frac{[\mu kR]}{[J]}$

Now $J$ is mechanical equivalent of heat. Since it converts heat into mechanical energy, it is dimensionless. Thus,

$[J]=1$

Also,

$[\mu kR]=[k][\mu R]=[S][S]=[S]^2$

Therefore,

$[\beta^2]=[S]^2 \quad\Rightarrow\quad [\beta]=[S]$

So,

$[\beta]=[S]=[k]=[\mu R]$

Hence option C is correct.

4. Find dimensions of $\alpha$

Since logarithm is dimensionless,

$[S]=[\alpha^2\beta]$

Using $[\beta]=[S]$,

$[S]=[\alpha^2][S]$

Thus,

$[\alpha^2]=1 \quad\Rightarrow\quad [\alpha]=1$

So $\alpha$ is dimensionless.

5. Check each option

Option A: $\alpha$ and $J$ have the same dimensions

We found:

$[\alpha]=1, \quad [J]=1$

So this is correct.

Option B: $S$ and $\alpha$ have different dimensions

We found:

$[S]=ML^2T^{-2}\Theta^{-1}, \quad [\alpha]=1$

So this is correct.

Option C: $S$, $\beta$, $k$ and $\mu R$ have the same dimensions

Yes, all are entropy-like quantities. So this is correct.

Option D: $\alpha$ and $k$ have the same dimensions

But

$[\alpha]=1, \quad [k]=[S]$

These are not the same. So this option is incorrect.

6. Final answer

The incorrect option is:

$\boxed{\text{D}}$