Do problems in P only reduce to NP and coNP problems?
One short answer, take $B$ be to be any problem in $\mathsf{P}$. The reduction is then powerful enough to simply solve the problem as part of the reduction, and consequently produce a trivial yes or no instance at the other end as appropriate. Then we can take $C$ and $D$ to be anything, they don't even have to be computable.

A more interesting, but conditional answer, is to take $B$, $C$ and $D$ to be any three $\mathsf{NP}$-complete problems (they don't even need to be different ones). Every $\mathsf{NP}$-complete problem is polynomial-time many-one reducible to every other $\mathsf{NP}$-complete problem, so the two reducibility conditions hold trivially. If $\mathsf{P} \neq \mathsf{NP}$, then $B$ can't be in $\mathsf{P}$, and if $\mathsf{NP} \neq co\mathsf{NP}$, then $C$ can't be in $co\mathsf{NP}$. You can do essentially the same thing by taking them all to be $co\mathsf{NP}$-complete. Taking it further, you can do this for any class in the polynomial hierarchy, and the statement being true would imply a collapse to whatever level you chose, another thing that is viewed as unlikely to happen. You could also choose a selection of $\mathsf{PSPACE}$-complete problems, and the statement would imply $\mathsf{P} = \mathsf{PSPACE}$, which also seems unlikely to be true.