Let's think about the countable case like this: think of the
binary tree $2^{\lt\omega}$, which has size $\omega$, but has
$2^\omega$ many branches. Each branch describes a cut in the
natural lexical order on the nodes, and so we have a countable
linear order with $2^\omega$ many cuts.

So consider a cardinal $\kappa$ and the tree $2^{\lt\kappa}$,
which admits a similar lexical order. If this tree has size
$\kappa$, then we get a linear order of size $\kappa$ with
$2^\kappa$ many cuts. And such a family of cuts turns into a chain
just as you point out in the question.

Thus, if $B$ has size $2^{\lt\kappa}$, then we can find $A$ of size
$2^\kappa$. In particular, whenever $2^{\lt\kappa}=\kappa$, then $P(\kappa)$
has a chain of size $2^\kappa$, as you desire.

In particular, under the GCH, the phenomenon will occur for every
infinite cardinal, since GCH implies $2^{\lt\kappa}=\kappa$ for
all infinite cardinals $\kappa$.

Lastly, let me point out that this argument method can be turned
into a characterization. Namely, the sets $B$ for which there is
an $A$ as you desire are exactly the sets $B$ of size $\kappa$ for
which there is a linear order on $\kappa$ with $2^\kappa$ many
cuts. The one direction we've established, and for the converse,
when there is such a chain $A$ in $P(B)$ of that size, then we may
place a pre-order on $B$ according to the order in which points
are added to sets in the chain (and extend this to a linear
order). Each element of $A$ gives a cut in this order, and so we
have a linear order on $\kappa$ with $2^\kappa$ many cuts.

So the phenomenon occurs for exactly those sets $B$ of size
$\kappa$ for which there is a linear order on $\kappa$ with
$2^\kappa$ many cuts.