We know that if we start with a ctm $\mathbb{B}$ and force with the poset of finite functions from $\omega$ to $2$, we add a single Cohen real. We also know that if we force with the poset $\mathbb{P} = Fn(\kappa \times \omega, 2, \aleph_0)$, we add $\kappa$ many reals (and hence can make the Continuuum Hypothesis fail).
What happens if instead we iterate adding one real $\kappa$ many times? Would we still get a model for not-CH?
Note that the definition of a Cohen forcing as $2^{<\omega}$ does not change between models.
Iterating it $\kappa$ many times, or taking the product of $\kappa$ many Cohen posets, or using $\mathbb P$ as you defined it -- all of these have the same consequence.
So to your question, yes. A finite-support iteration of length $\kappa$ of adding a single Cohen at a time would end up with a model of $\lnot$CH.