Equivalence definition matters here. But, I-equivalence is a common query in Bayesian Networks, in which I just assumed you're asking for that, but you need to specify in your question as well. For two BNs to be I-equivalent, they need to have the same skeleton, and immoralities. Check the lecture here. Immorality is defined as a V-structure that a node $X$, having parents $Y,Z$ without an edge in between them.
In your graphs, the skeletons are clearly the same. And, the only V-structure in both graphs is the $B,C,D$ triple. So, the two graphs have the same V-structures and they are I-equivalent.
You can read section 9 of this article and this piece of information would be helpful:
I-equivalence can be determined by looking at the skeleton graphs, i.e. the graphs formed by converting directed edges to undirected edges. Two graphs are I-equivalent if they have the same skeleton and the same set of v-structures
In your case, you just compare the v-structures in each graph(both contain only $B\rightarrow C \leftarrow D$) and remove the direction to compare the skeleton(the same), so they are I-equivalent.