If $M$ stops in no more than 50 steps, than the positions $M$ can reach on the normally infinite tape are limited. Thus the infinite tape can be simulated by a finite one. This means that the tape can be simulated by a finite automaton. It follows that a turing machine $M$ that stops in no more than 50 steps is bisimilar to some finite automaton $M'$.
Let $Q$ be the set of states of $M$, $F \subset Q$ the set of accepting states and $\Gamma$ be the alphabet.
Then we build the set of states $Q'$ of $M'$ as follows:
$Q' = \{ \langle n, q, s, p, a\rangle \, | n \in \{0,...,50\} \, q \in Q, s \in \Gamma, p \in \{-50,...,50\}, a \equiv q \in F\}$ where $p$ is the position of the read/write head above the tape. We can restrict the position to $\{-50,...,50\}$ because the number of allowed computing steps limits the number of reachable positions.
Having a state $\langle n, q, s, p, a\rangle$ of the finite automaton $M'$ then means that we are at state $q$ of the original automaton, with $s$ on the tape at position $p$ where also the read/write head is positioned, after the $n$-th computing step. The state is an accepting one if $a \equiv true$.
Transforming the transition relation of a concrete turing machine is a little more work but not necessary for the original question, because it is enough to show that the state space is finite (and thus we can just test each input with a length of at most 50 symbols on each such automaton). The idea is to build a new transition relation that goes from a state $\langle n, q, s, p, a \rangle$ to a state $\langle n+1, q', s', p', a'\rangle$ in the $n$-th computing step iff the transition $\langle q, s, p\rangle \rightarrow \langle q', s', p'\rangle$ was in the original transition relation.