Let $C \ge 1$ be the capacity of the knapsack and let $G_1, \dots,G_k$ be your groups, where each group is a non-empty collection of items, i.e., $G_i = \{ x^{(i)}_1, x^{(i)}_2, \dots \}$. Groups are pairwise disjoint. Define $n= \sum_{i=1}^k |G_i|$ as the total number of items in the instance.
For any item $x$, let $w(x)$ and $v(x)$ denote its weight and value, respectively, and assume for simplicity that the all weights and the values of the above items positive. It is helpful to imagine the existence of $k$ additional items $x_1^{(0)}, \dots, x_k^{(0)}$, each with value and weight $0$.
Here are three possible algorithms. Pick the fastest one according to your instance.
For $i=0, \dots, k$ and $c=0, \dots, C$, define $OPT[i, c]$ as the maximum overall value that can be obtained by selecting at most one element from each of the group $G_1, \dots, G_i$ and no elements from the groups $G_{i+1}, \dots, G_k$, with the constraint that the total weight of the selected elements must be at most $c$.
According the above definition we have $OPT[0,c] = 0$ and, for $i > 0$
$$
OPT[i,c] = \max_{\substack{j=0, \dots, |G_i| \\ w(x^{(i)}_j) \le c}} \left( v(x^{(i)}_j) + OPT\left[i-1, c-w(x^{(i)}_j)\right] \right).
$$
You can compute all the quantities $OPT[i,c]$ in non-decreasing order of $i$. Since $OPT[i,c]$ can be fond in time $O(|G_i|)$, the overall time complexity is
$$
O\left( C \cdot \sum_{i=1}^k |G_i| \right) = O(C \cdot n).
$$
The value of the optimal solution is $OPT[k, C]$ and an optimal set of items to select can be found using standard techniques.
Let $V$ be an upper bound on the maximum attainable value. For example: $V = \sum_{i=1}^k \max G_i$.
For $i=0, \dots, k$ and $v=0, \dots, V$, define $OPT[i, v]$ as the minimum overall weight of a set of items that has an overall value of at least $v$ and can be obtained by selecting at most one element from each of the group $G_1, \dots, G_i$ and no elements from the groups $G_{i+1}, \dots, G_k$.
If no such set of items exists, define $OPT[i, v]= +\infty$.
We have $OPT[0, 0] = 0$, and $OPT[0, v] = +\infty$ for $v>0$. Moreover, for $i>0$:
$$
OPT[i, v] = \min_{j=0,\dots,|G_i| } \left( w(x^{(i)}_j) + OPT\left[i-1, \max\{v - v(x^{(i)}_j), 0 \}\right] \right).
$$
The value of the optimal solution is $\max \{v \mid v=0, \dots, V \wedge OPT[k, v] \le C\}$.
Again, you can solve all subproblems in non-decreasing order of $i$, where each $OPT[i, v]$ can be found in time $O(|G_i|)$ yielding an overall time complexity of $O(V \cdot n)$.
Partition the groups into three sets $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3$ such that $\sum_{G_i \in \mathcal{G}_1} |G_i| \le n/2$, $\sum_{G_i \in \mathcal{G}_2} |G_i| \le n/2$, and $|\mathcal{G}_3| = 1$.
Guess an element $x^*$ that belongs to the only group in $\mathcal{G}_3$. There are at most $O(n)$ choices.
Exhaustively generate all sets that can be constructed by selecting at most one element from each group in $\mathcal{G}_1$ (and no elements from other groups). For each such set $S$ look at the pair $(\sum_{x \in S} w(x), \sum_{x \in S} v(x))$ and collect all these pairs into a list $L_1$. Notice how there are at most $O(2^{n/2})$ sets to examine, which means that we can build $L_1$ in time $O(n \cdot 2^{n/2})$.
Repeat the same procedure for $\mathcal{G}_2$ to obtain a list $L_2$. Sort the pairs $(w,v)$ in $L_2$ in non-decreasing order of $w$ and use this sorted list to compute a value $\nu(w)$ for each $w$ that appears in some pair $(w, v)$ in $L_2$. Specifically, $\nu(w) = \max_{\substack{(w',v') \in L_2 \\ w' \le w}} v'$, i.e., $\nu(w)$ is the maximum value that can be obtained by selecting at most one item from each set in $\mathcal{G}_2$ without exceeding an overall weight of $w$.
Sorting $L_2$ requires time $O(n \cdot 2^{n/2})$, while the quantities $\nu(w)$ can be found by a linear scan of (the sorted version) of $L_2$.
Finally, for each pair $(w, v)$ in $L_1$ with $C-w-w(x^*) \ge 0$, we can fetch $\nu(C-w-w(x^*))$ in time $O(\log |L_2|) = O(\log 2^{n/2}) = O(n)$ and consider the quantity $v + v(x^*) + \nu(v)$ as a candidate answer.
The maximum among all candidate answers is the value of an optimal solutions to your original instance. The overall time spent is $O(n^2 \cdot 2^{n/2})$.
