I believe that it must split if $C$ is abelian.
More precisely, I will prove that if $C$ is an abelian $p$-group and $D$ is a $p'$-group, then $C$ has a complement in $G$ and so $G \cong C \rtimes (D \rtimes B)$. We are assuming that $G$ is finite. Note that under these assumptions $C$ is a characteristic subgroup of $A$, and hence normal in $G$.
There is a general result that says that, if $Q$ is a $p'$-group acting on a finite abelian $p$-group $P$, then $P = C_P(Q) \times [P,Q]$. I don't have a reference to hand, but I know it is in Gorenstein's book on Finite Groups, and I can find the precise reference later. Note that this does not hold in general for nonabelian $p$-groups.
So we have $C=C_C(D) \times [C,D]$.
Now $A/[C,D]$ is the direct product $C/[C,D] \times D[C,D]/[C,D]$. Since $A/D[C,D] \cong C/[C,D]$ is abelian, we have $[A,A] \le D[C,D]$ and so $[C,D] = C \cap [A,A]$ is normal in $G$.
Since the direct factors $D[C,D]/[C,D]$ and $C/[C,D]$ of $A/[C,D]$ have coprime orders, they are both characteristic in $A/[C,D]$ and hence normal in $G/[C,D]$. In particular, $D[C,D]/[C,D]$ is normalized by the subgroup $B[C,D]/[C,D]$, which is a complement of $A/[C,D]$ in $G/[C,D]$.
Now the subgroup $BD[C,D]$ of $G$ has contains $B$ and $D$ and intersects $C$ in $[C,D]$. So it satisfies the original hypotheses, but with $C$ replaced by $[C,D]$. So if we now replace $G$ by $BD[C,D]$, we have $C_C(D)=1$.
Now let $N = N_G(D)$. Since, by the Schur-Zassenhaus Theorem, all complements of $C$ in $A$ are conjugate (to $D$), the Frattini Argument shows that $G=AN$,
and hence, since $D \le N$, $G=CN$. Also $[C \cap N,D] \le C \cap D = 1$, so $C \cap N \le C_C(D) = 1$, and hence $C \cap N = 1$. So $N$ is a complement of $C$ in $G$. (It is also a complement of the original $C$ that we replaced by $[C,D]$.)
Note that this argument doesn't use the facts that $D$ and $B$ are cyclic, but it does use the assumption that $C$ and $D$ have coprime orders.
I do not know whether such an extension always splits when $C$ is allowed to be a nonabelian $p$-group.
Best Answer
No.
Let $X$, $Y$, and $Z$ be infinite cyclic groups with generators $x,y,z$. Make a semidirect product $XY$ using the nontrivial action of $X$ on $Y$. Make the direct product of this with $Z$. In this group there is the free abelian group $XZ$, and inside that there is the infinite cyclic group generated by $xz$. This has two "normal complements" $XY$ and $YZ$. One is nonabelian and the other is abelian.