On p. 226 of his ${\it Set}$ ${\it Theory}$, Jech considers adding $\lambda$ many subsets of $\kappa$ to a ground model $M$. He outlines a suitable partial order on the assumption that $M$ satisfies $\kappa<\lambda$, $2^{<\kappa} = \kappa$, and $\lambda^\kappa = \lambda$. Any resulting generic extension, he claims, will preserve cardinals and satisfy $2^\kappa =\lambda$. $\DeclareMathOperator{\dom}{dom}\DeclareMathOperator{\rng}{rng}$
It looks like we can get a generic extension with the same properties without assuming $2^{<\kappa} = \kappa$ (though we do assume that $\kappa<\lambda$ and $\lambda^\kappa = \lambda$). In particular, let $P$ be the set of all $p$ such that:
(i) $\dom(p)\subseteq \lambda\times\kappa$ and $|\dom(p)|<\omega$
(ii) $\rng(p)\subseteq\{0,1\}$
and let $p$ be stronger than $q$ iff $p\supseteq q$.
Since $P$ is c.c.c., cardinals will be preserved. In addition, we seem to have:
$$(2^\kappa)^{M[G]} \leq (|B|^\kappa)^{M} \leq (|P|^\kappa)^{M}\leq (\lambda^\kappa)^{M} \leq \lambda \leq (2^\kappa)^{M[G]}$$
So, $\lambda = (2^\kappa)^{M[G]}$.
Am I missing something?
Your argument is fine but you should consider the other effects your forcing has. Since you take finite conditions, new small sets will appear in the extension. In particular, there will be $\lambda$ many new reals there (in fact your forcing is just isomorphic to adding $\lambda$ many Cohen reals). So the fact that you blew up $2^\kappa$ is really because you blew up all the power sets below that as well. This doesn't happen with Jech's forcing, which makes it better in a sense since it allows for finer control of the extension.