If Pavlik knows the product, he could determine the product of the last two numbers that were not chosen. Since he could not determine the pairing of the product, then he could not determine the numbers that were not chosen although he knows their product. Consider all products of two numbers in the given set:
$$\begin{aligned} 1 \cdot 2 &= 2, \quad 1 \cdot 3 = 3, \quad 1 \cdot 4 = 4, \quad 1 \cdot 5 = 5, \quad 1 \cdot 6 = 6, \quad 1 \cdot 7 = 7, \quad 2 \cdot 3 = 6, \quad 2 \cdot 4 = 8, \\ 2 \cdot 5 &= 10, \quad 2 \cdot 6 = 12, \quad 2 \cdot 7 = 14, \quad 3 \cdot 4 = 12, \quad 3 \cdot 5 = 15, \quad 3 \cdot 6 = 18, \quad 3 \cdot 7 = 21, \\ 4 \cdot 5 &= 20, \quad 4 \cdot 6 = 24, \quad 4 \cdot 7 = 28, \quad 5 \cdot 6 = 30, \quad 5 \cdot 7 = 35, \quad 6 \cdot 7 = 42. \end{aligned}$$
There are only 2 products that appear more than once. This is $1 \cdot 6 = 6 = 2 \cdot 3$ and $2 \cdot 6 = 12 = 3 \cdot 4$. Suppose that the product of the two numbers that were not chosen equals $6$. In both cases the sum of the numbers is odd: $1+6=7$ and $2+3=5$. Thus the sum of chosen numbers is also odd. This contradicts Pavlik's claim. Therefore, the product of the two numbers that were not chosen might equal $12$. In this case their sum might be even ($2+6=8$) or odd ($3+4=7$). Then the product of 5 chosen numbers is equal to $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{12} = 420$.