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 $(\tilde B_s)_{s\in[0,t]}:=(B_s-B_t)_{s\in[0,t]}$ 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.