The permutation matrix $P$ corresponding to the permutation $$ \sigma \colon \begin{cases} i \mapsto 2i & \text{ if } i \in [1,n] \\ i \mapsto 2i - 2n - 1 & \text{ if } i \in [n+1,2n] \end{cases} $$ does the job.
Edit: As requested, I am providing some details. Notice that the original matrix $B$ has zero diagonal, which implies that $C$ also has zero diagonal, i.e., it has a non-zero subdiagonal and a non-zero superdiagonal but zeros everywhere else. There are precisely $4n-2$ non-zero entries in $B$ and also precisely $4n-2$ non-zero entries in the matrix $C$ we are looking for. We have (for $n=4$, for example) $$ B = \begin{pmatrix} &&&& \color{red}{*} & \color{red}{*} && \\ &&&&& \color{red}{*} & \color{red}{*} & \\ &&&&&& \color{red}{*} & \color{red}{*} \\ &&&&&&& \color{red}{*} \\ \color{blue}{*} &&&&&&& \\ \color{blue}{*} & \color{blue}{*} &&&&&& \\ & \color{blue}{*} & \color{blue}{*} &&&&& \\ && \color{blue}{*} & \color{blue}{*} &&&& \end{pmatrix}$$ and the permutation $\sigma$ transforms $B$ into the matrix $$ C = \begin{pmatrix} & \color{blue}{*} &&&&&& \\ \color{red}{*} && \color{red}{*} &&&&& \\ & \color{blue}{*} && \color{blue}{*} &&&& \\ && \color{red}{*} && \color{red}{*} &&& \\ &&& \color{blue}{*} && \color{blue}{*} && \\ &&&& \color{red}{*} && \color{red}{*} & \\ &&&&& \color{blue}{*} && \color{blue}{*} \\ &&&&&& \color{red}{*} & \\ \end{pmatrix}.$$