MIP is already exact. Let there be

- $l = 26$ letters;
- $m = 5$ maximum group items;
- $n = 7$ as an example item count, also equal to the maximum number of groups;
- $G_{i,j} \in \lbrace 0,1 \rbrace$ binary group-item assignments, $0 \le i < n$, $0 \le j < n$;
- $I_{j,k} \in \lbrace 0,1 \rbrace$ binary item-letter assignments, $0 \le j < n$, $0 \le k < l$;
- $L_{i,k}\in \lbrace 0,1 \rbrace$ binary group-letter assignments, $0 \le i < n$, $0 \le k < l$.

Minimize
$$\sum_i \sum_k L_{i,k}$$
subject to
$$1=\sum_i G_{i,j} \;\forall j$$
$$m \ge \sum_j G_{i,j} \;\forall i$$
$$n L_{i,k} \ge \sum_j I_{j,k} G_{i,j} \;\forall i,k$$

Works fine:

```python
from string import ascii_uppercase
import scipy.sparse as sp
import numpy as np
from scipy.optimize import milp, Bounds, LinearConstraint


items = (
    'ABC', 'CDF', 'EHI', 'AHP', 'EPZ', 'CIZ', 'ITW',
)

L = len(ascii_uppercase)
M = 5  # max items per group
N = len(items)


# Variables: NN item-group assignments; NL letter-group assignments
cost = np.concatenate((
    np.zeros(N*N),  # group-item assignments non-optimized
    np.ones(N*L),   # group-letter assignments minimized
))

# Every item must be assigned to exactly one group
item_excl = LinearConstraint(
    A=sp.hstack(
        (sp.eye(N),)*N + (sp.csc_matrix((N, N*L)),),
        format='csc',
    ),
    lb=np.ones(N), ub=np.ones(N),
)

# Every group can have at most M items (Kronecker)
group_capacity = LinearConstraint(
    A=sp.hstack((
        sp.kron(sp.eye(N), np.ones(N)),
        sp.csc_matrix((N, N*L)),
    ), format='csc'),
    ub=np.full(shape=N, fill_value=M),
)

# For each letter of each item, that item must have at least one group assignment that in turn has
# a compatible letter assignment
item_block = sp.dok_array((L, N))
letter_block = sp.dok_array((L, L))
n_letters = 0
for letter in ascii_uppercase:
    items_with_letter = [i for i, word in enumerate(items) if letter in word]
    if len(items_with_letter) > 0:
        item_block[n_letters, items_with_letter] = -1
        letter_block[n_letters, ord(letter) - ord('A')] = N
        n_letters += 1
item_block = sp.block_diag((item_block[:n_letters, :],)*N, format='csc')
letter_block = sp.block_diag((letter_block[:n_letters, :],)*N, format='csc')

letter_assigns = LinearConstraint(
    A=sp.hstack((item_block, letter_block), format='csc'),
    lb=np.zeros(n_letters*N),
)

result = milp(
    c=cost,
    integrality=np.ones(N*N + N*L),  # all variables integral
    bounds=Bounds(lb=0, ub=1),       # all variables binary
    constraints=(item_excl, group_capacity, letter_assigns),
)
assert result.success, result.message

item_assign, letter_assign = np.split(result.x, (N*N,))
item_assign = item_assign.reshape((N, N)).round().astype(int)
letter_assign = letter_assign.reshape((N, L)).round().astype(int)

print('Item assignments:')
print(item_assign, end='\n\n')

print('Letter assignments:')
print(letter_assign, end='\n\n')

for i_group in range(N):
    group_assigns = item_assign[i_group]
    if group_assigns.any():
        group_letters = ''.join(ascii_uppercase[i] for i in letter_assign[i_group].nonzero()[0])
        print(f'Group {i_group} has letters {group_letters}, items',
              ', '.join(items[i] for i in group_assigns.nonzero()[0]))
```

```none
Item assignments:
[[1 0 1 1 1 1 0]
 [0 0 0 0 0 0 1]
 [0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0]
 [0 1 0 0 0 0 0]]

Letter assignments:
[[1 1 1 0 1 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1]
 [0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]

Group 0 has letters ABCEHIPZ, items ABC, EHI, AHP, EPZ, CIZ
Group 1 has letters ITW, items ITW
Group 6 has letters CDF, items CDF
```