Is a DTM with k-tapes not the same thing as a NDTM with k-branches?
Let $\mathsf{P^k}$ be the class of problems solvable in polynomial time on deterministic Turing machines with $k$ tapes. Then
$$
\mathsf{P^1} = \mathsf{P^2} = \mathsf{P^3} = \mathsf{P^4} = \mathsf{P^5} = \cdots
$$
Let $\mathsf{PD^{d}}$ be the class of problems solvable in polynomial time on deterministic Turing machines with access to a $d$-dimensional infinite tape. Then
$$
\mathsf{P^1} = \mathsf{PD^1} = \mathsf{PD^2} = \mathsf{PD^3} = \mathsf{PD^4} = \mathsf{PD^5} = \cdots
$$
Similarly, $\mathsf{P}$ is also the class of problems solvable in polynomial time on RAM machines, which have a random-access memory (i.e. you can implement indexing $A[i]$ without needing the head to scan to the $i$th position of the tape).

The definition of $\mathsf{P}$ is very robust to the particular machine model used, and this makes $\mathsf{P}$ a very natural complexity class. This is similar to the class of all decidable languages, which can be defined in countless ways. The exact way you define $\mathsf{P}$ is arbitrary, since all definitions amount to the same complexity class.

The story gets a bit more complicated when we consider computation models other than deterministic machines (of any kind). For example, are probabilistic polytime algorithms stronger than deterministic ones? The answer might depend on the error. Probabilistic polytime algorithms with constant error probability define the complexity class $\mathsf{BPP}$, which is believed (by most) to be the same as $\mathsf{P}$. If we allow unbounded error we get $\mathsf{PP}$, which contains $\mathsf{NP}$, and so such algorithms are probably more powerful than deterministic ones. Similarly, quantum algorithms with bounded errors form the complexity class $\mathsf{BQP}$, which contains problems such as factoring which are not believed (by most) to be solvable in classical polynomial time.