It is true that $\mathbb{co} A = B$, but the proof is a little tedious.
Since $A \subset B$, it is clear that $\mathbb{co} A \subset B$. The other direction requires a little more work:
Both $A$ and $B$ are closed bounded sets, hence compact. $B$ is convex. The convex hull $\mathbb{co} A$ is closed and compact (follows from compactness and Carathéodory's theorem).
Let $E$ be the set of extreme points of $B$. The basic idea is to show that $E \subset A$. Since the Krein-Milman theorem tells us that $B = \overline{\mathbb{co}} E$, it then will follow that $B \subset \mathbb{co} A$.
To characterize the extreme points, I need some notation and a lemma.
Let $C = \{ x \in \mathbb{R}^n | \; g_i (x) \leq 0, \; \forall i \in I \}$, where each $g_i$ has the form $g_i(x) = \langle \gamma_i, x \rangle - \beta_i$. (Note that $B$ has this form.) Define the 'active index set' $I_{x_0} = \{ i \in I | g_i(x_0) = 0 \}$.
Lemma: $x_0$ is an extreme point of $C$ iff $x_0 \in C$ and $\mathbb{sp} \{ \gamma_i \}_{i \in I_{x_0}} = \mathbb{R}^n$.
Proof: ($\implies$): Suppose $x_0$ is an extreme point of $C$, but $\mathbb{sp} \{ \gamma_i \}_{i \in I_{x_0}} \neq \mathbb{R}^n$. Then choose a non-zero $h \in \mathbb{sp} \{ \gamma_i \}_{i \in I_{x_0}}^{\bot}$. Furthermore note that if $ i \notin I_{x_0}$, then there exists a neighborhood $U$ of $x_0$ such that $i \notin I_x$, forall $x \in U$. Now consider the points $x_0 + \lambda h$, and observe that for $\lambda$ small enough, $x_0 \pm\lambda h \in C$ (since all constraints are satisfied). We can write $x_0 = \frac{1}{2} ( (x_0+\lambda h) + (x_0-\lambda h))$ which contradicts the fact that $x_0$ is extreme.
($\impliedby$): Now suppose $x_0 \in C$ and $\mathbb{sp} \{ \gamma_i \}_{i \in I_{x_0}} = \mathbb{R}^n$. Suppose $x_0 = \lambda x + (1-\lambda) y$, with $x,y \in C$, and $\lambda \in (0,1)$. Let $i \in I_{x_0}$, then $g_i(x_0) = \lambda g_i(x) + (1-\lambda) g_i(y) = 0$. It follows that $g_i(x) = g_i(y) = 0$, from which we have $\langle \gamma_i, x_0 \rangle = \langle \gamma_i, x \rangle = \langle \gamma_i, y \rangle$, for all $i \in I_{x_0}$. However, this implies that $x_0-x$ and $x_0-y$ lie in the complement of $\mathbb{sp} \{ \gamma_i \}_{i \in I_{x_0}}$, which is $\{ 0 \}$ by assumption. Hence $x = y = x_0$. It follows that $x_0$ is extreme.
Now back to the set $B$. Let $x$ be an extreme point of $B$. $B$ is defined by $2n+1$ constraints, however of the $2n$ constraints $x_i \in [0,1]$, at most $n$ can be active. Hence at most $n+1$ constraints can be active (and any $n$ of these are linearly independent).
There are two possibilities: (1) $x_1+...+x_n = k$. In this case, $n-1$ of the other constraints must be active (ie, $x_i$ is either $0$ or $1$ for $n-1$ indices $i$). Since $k$ is an integer, it then follows that the remaining $x_i$ must also be $0$ or $1$. Thus $x = \sum_{j \in J} e_j$, where $|J| = k$. (2) $x_1+...+x_n < k$. In this case all of the $x_i$ must be either $0$ or $1$ (at most $k-1$ can be $1$, of course). This gives $x = \sum_{j \in J} e_j$, where $|J| < k$.
This characterizes the extreme points of $B$ as having the form $x = \sum_{j \in J} e_j$, where $|J| \leq k$, and it a straightforward observation to note that all such $x$ satisfy $x \in A$. Hence we have the desired result.