Assume $u_{n+1}=Pu_n$, then $M_k\colon=u_{n-k}(X_k)$, $k=0,\ldots,n$ is a martingale.
By the Markov property and $Pu_{n-k-1}=u_{n-k}$ we get $$ \E^x(M_{k+1}|\F_k) =\E^x(u_{n-k-1}(X_{k+1})|\F_k) =Pu_{n-k-1}(X_k) =u_{n-k}(X_k) =M_n~. $$