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.