The total number of pairs of shoes is 20, and the number of pairs with designer labels is 8. Therefore, the probability of choosing a pair with designer labels is \(p = \frac{8}{20} = 0.4\).

The probability of choosing a pair without designer labels is \(1 - p = 0.6\).

We need to find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of 7. This is a binomial probability problem where \(n = 7\), \(p = 0.4\), and we want \(P(X \leq 4)\).

Using the complement rule, \(P(X \leq 4) = 1 - P(X = 5) - P(X = 6) - P(X = 7)\).

Calculate each probability:

\(P(X = 5) = \binom{7}{5} (0.4)^5 (0.6)^2 = 21 \times 0.01024 \times 0.36 = 0.0774144\).

\(P(X = 6) = \binom{7}{6} (0.4)^6 (0.6)^1 = 7 \times 0.004096 \times 0.6 = 0.0172032\).

\(P(X = 7) = \binom{7}{7} (0.4)^7 (0.6)^0 = 1 \times 0.0016384 \times 1 = 0.0016384\).

Thus, \(P(X \leq 4) = 1 - (0.0774144 + 0.0172032 + 0.0016384) = 1 - 0.096256 = 0.903744\).

Rounding to three decimal places, the probability is \(0.904\).