The given relation $R(A,B,C,D)$ and $FDs$ of $R$ = $\left \{ A \rightarrow B, B \rightarrow C, C \rightarrow D, D \rightarrow B \right \}$
So, We have the following results from the FD set of $R$ :
$A \rightarrow B,C,D$
$B \rightarrow C,D$
$C \rightarrow D,B$
$D \rightarrow B,C$
1. Dependency Preservation :
We say that in any decomposition {$R1,R2...Rn$} of $R$, Dependency is Preserved when $\left \{ F1 \cup F2 \cup F3....\cup Fn \right \}$ covers $F$ where $F$ is the FD set of $R$ and $Fn$ is the FD set of $Rn$.
So, here in given question, $F1 = \left \{ A \rightarrow B \right \}$
$F2 = \left \{ B \rightarrow C, C \rightarrow B \right \}$
$F3 = \left \{ B \rightarrow D, D \rightarrow B \right \}$
$F1 \cup F2 \cup F3 = \left \{ A \rightarrow B, B \rightarrow C, C \rightarrow B, B \rightarrow D, D \rightarrow B \right \}$
We can see that $F1 \cup F2 \cup F3$ Covers $F$. Hence, Dependency Preserved.
2. Lossless Join :
The strategy to find Lossless Join or Not is that Try to merge all $Ri$ where $i = 1 \,\,to\,\, n$ by taking Two Relations at a time such that the common attribute(s) is a Key in at least one of the tables being merged. If there is some way(at least one) of doing so then Decomposition is Lossless.
Here in the Question, We can merge $R1 \,\,and \,\, R2$ because the common attribute ($B$) is a key in one of the tables ($R2$) and after that We can merge $R3$ with them because then also the common attribute ($B$) is a key in at least one of the tables. So, The decomposition of $R$ into $R1,R2,R3$ is Lossless Join decomposition.