Assume that $\kappa([A, B])$ is the condition number of a block matrix $[A, B]$. Given that, we also know, $$\kappa(C) < \kappa(A)$$
I am curious whether if the following assertion is true or when does that inequality may hold: $$\kappa([C,B]) < \kappa([A, B])$$
Both $A$ and $C$ are $n \times n$ symmetric square matrices and B is an $n\times m$ rectangular matrix.
The condition number of a rectangular matrix $B$ is defined as the ratio of largest and smallest nonzero singular values,
$$\kappa(B) = \frac{\sigma_{\max}(B)}{\sigma_{\min}(B)}$$
where $\sigma_{\max}(B)$ is the largest singular value of $B$ and $\sigma_{\min}(B)$ is the smallest nonzero singular value of matrix $B$.
Your question is not very clear.
If the $A^\ast$ in your question simply denotes some other matrix than $A$ (rather than the conjugate transpose), you may consider $$A_0=\pmatrix{3\\ &3},\ A=\pmatrix{4\\ &3},\ B=\pmatrix{0\\ 4}.$$ We have $\kappa(A_0)=1<\frac43=\kappa(A)$ but $\kappa([A_0,B])=\frac53>\frac54=\kappa([A,B])$.
If $A^\ast$ does mean the conjugate transpose of $A$ (i.e. $\bar{A}^T$), then your statement that $\kappa(A^\ast)<\kappa(A)$ is always false because a matrix and its conjugate transpose always have identical singular values. Nevertheless, you may still ask if there exist $A$ and $B$ such that $\kappa([\bar{A}^T,B])$ is larger than or smaller than $\kappa([A,B])$. Since $\kappa([\bar{A}^T,B])$ and $\kappa([A,B])$ in practice are seldom the same (unless $A$ is normal, as pointed out by user251257 in a comment), if we interchange the roles of $A$ and $\bar{A}^T$ when necessary, we can almost always construct an example of $\kappa([\bar{A}^T,B])>\kappa([A,B])$ or $\kappa([\bar{A}^T,B])<\kappa([A,B])$ at will out of random samples of $A$ and $B$.