Use dimensional consistency in $AD=C\ln(BD).$

Since the argument of logarithm must be dimensionless, $[BD]=1 \implies [B][D]=1 \implies [B]=[D]^{-1}.$

Also, since $\ln(BD)$ is dimensionless, the RHS has dimensions of $C$ only. Hence $[AD]=[C] \implies [A][D]=[C].$

Now check each option.

Option A: $A^2-B^2C^2$

For subtraction, dimensions must be same.

$[A^2]=[A]^2$ and $[B^2C^2]=[B]^2[C]^2=([B][C])^2.$

But $[C]=[A][D], \quad [B]=[D]^{-1}$ so $[B][C]=[D]^{-1}[A][D]=[A].$ Therefore, $[B^2C^2]=[A]^2=[A^2].$

So option A is meaningful.

Option B: $\frac{A-C}{D}$

For $A-C$ to be meaningful, we need $[A]=[C]$. But from $[C]=[A][D],$ this would require $[D]=1,$ which is impossible because none of the quantities is dimensionless.

Hence $A-C$ itself is not meaningful.

So option B is not meaningful.

Option C: $\frac{A}{B}-C$

Check dimensions: $\left[\frac{A}{B}\right]=[A][B]^{-1}=[A][D].$ Since $[C]=[A][D]$, $\left[\frac{A}{B}\right]=[C].$

Therefore subtraction is valid. Option C is meaningful.

Option D: $\frac{C}{BD}-\frac{AD^2}{C}$

First term: $[BD]=1 \implies \left[\frac{C}{BD}\right]=[C].$

Second term: $\left[\frac{AD^2}{C}\right]=\frac{[A][D]^2}{[C]}.$ Using $[C]=[A][D]$, $\left[\frac{AD^2}{C}\right]=\frac{[A][D]^2}{[A][D]}=[D].$

Thus the two terms have dimensions $[C]$ and $[D]$, which are different (given all quantities have different dimensions).

Hence subtraction is invalid. Option D is not meaningful.

Final result

The combinations that are not meaningful are: $\boxed{\text{B, D}}$