Solution:

Let $P$ be said product. Evidently there are two distinct sets of numbers $\{x_{1}, x_{2}, x_{3}, x_{4}\}$ and $\{y_{1}, y_{2}, y_{3}, y_{4}\}$ such that $x_{1} x_{2} x_{3} x_{4} = y_{1} y_{2} y_{3} y_{4} = P$, but $x_{1} + x_{2} + x_{3} + x_{4}$ and $y_{1} + y_{2} + y_{3} + y_{4}$ have different parity.

Instead of considering the four numbers Alice picks, we consider instead the two numbers Alice does not pick; $\{x_{5}, x_{6}\}$ and $\{y_{5}, y_{6}\}$. They have the same properties we described above, since

$$x_{5} x_{6} = \frac{6!}{x_{1} x_{2} x_{3} x_{4}} = \frac{720}{P} = \frac{6!}{y_{1} y_{2} y_{3} y_{4}} = y_{5} y_{6}$$
and analogously the sums also have different parities.

Thus we are looking for pairs of distinct numbers in $\{1,2,3,4,5,6\}$ which have the same product but different sums. We can record the entire multiplication table, as below.

Of these, 12 has the desired property but 6 does not. Hence, Alice chose one of the quadruples $\{1,3,4,5\}$ or $\{1,2,5,6\}$; the product is then $P = 1 \cdot 3 \cdot 4 \cdot 5 = 1 \cdot 2 \cdot 5 \cdot 6 = 60$.