Here on wikipedia is claimed that the process $X_t:=\sup_{s \in [0,t]} B_s-B_t$ is distributed like $\vert B_t \rvert$ where $B_t$ is standard Brownian motion.
On the other hand, it is claimed here in Corollary $6.21$ that $\sup_{s \in [0,t]} B_s$ is distributed like $\vert B_t \rvert.$
So how is it possible that $\sup_{s \in [0,t]} B_s-B_t$ is distributed like $\sup_{s \in [0,t]} B_s.$ There seems to be something wrong with probability.
If you have any further questions, please let me know.
To conclude that something here is wrong, you should be able to indicate a contradiction. Of course, you won't be able to do this, since the two random variables, $M_t:=\sup_{s \in [0,t]} B_s$ and $M_t-B_t$, have indeed the same distribution, for each real $t>0$.
To quickly see this, note that $(\tilde B_s)_{s\in[0,t]}:=(B_{t-s}-B_t)_{s\in[0,t]}$ is a standard Brownian motion on the interval $[0,t]$ and hence equals $(B_s)_{s\in[0,t]}$ in distribution. So, $\tilde M_t:=\sup_{s \in [0,t]} \tilde B_s=M_t-B_t$ equals $M_t$ in distribution.
Perhaps it would be a bit easier to comprehend the following discrete version of the same result: Let $X_1,X_2,\dots$ be iid random variables each taking each of the two values $a,-a$ with probability $1/2$, for some real $a>0$. Let $M_n:=\max_{0\le j\le n}S_j$, where $S_j:=X_1+\dots+X_j$ (as usual, let the sum of any empty family equal $0$). Note that $S_{n-j}-S_n=\tilde S_j:=\tilde X_1+\dots+\tilde X_j$ for each $j=0,\dots,n$, where $\tilde X_i:=-X_{n+1-i}$. Since $(\tilde X_i)_{i=1}^n$ equals $(X_i)_{i=1}^n$ in distribution, we see that $(\tilde S_j)_{j=0}^n$ equals $(S_j)_{j=0}^n$ in distribution, whence \begin{equation} M_n-S_n=\max_{0\le j\le n}(S_j-S_n)=\max_{0\le j\le n}(S_{n-j}-S_n)=\max_{0\le j\le n}\tilde S_j \end{equation} equals $M_n$ in distribution.
