Finding the Length of the shortest Accepting path of a NDTM Let $M$ be a NDTM (non deterministic Turing machine) which decides a certain NP-complete language, say SAT.

$M$ computes any instance $I$ of the NP-complete problem in at most $p(n)$ non deterministic steps ($p$ in a polynomial function, $n$ is the input size). The length of a computation path can be measured as the "rank" of the non deterministic step where $M$ halts. Let call $rank_M(I)$  the rank of the shortest accepting path ($rank_M(I)$ is $p(n)+1$ if $M$ rejects $I$).

Is this problem NP-complete : Given a boolean expression $I$, is it true that $rank_M(I)=k$ ? ($k$ is polynomially bounded in $n$)

Edit --- $rank_M(I)$ has been modified : it defines now the shortest accepting path (instead of just the rank of the step where $M$ halts).

Thank you.
Your problem is $\mathsf{NP}$-complete, and you can find a proof for instance here (I gave a sketch in comments). 

A remark: If you consider $\{\langle M,x,t\rangle : M$ halts on $x$ in $t$ steps $\}$, then you can only show it is $\mathsf{NP}$-hard. But as you mention that you consider $t$ polynomially bounded (in the sizes of $M$ and $x$, I guess), then your problem belongs to $\mathsf{NP}$.