There are 278400 ways to assign the people.

Note that each family can be partitioned in either a (4, 1) fashion (where exactly one parent sits with three children and the other is alone) or a (3, 2) fashion (where one parent sits with two children and the other sits with one).

Case 1. All families are partitioned in (4, 1) fashion.
The 4-group of each family occupies a table, and the other four parents take the remaining table. There are 2 ways to choose a parent of each family. So the number of ways is $2^4 = 16$ if we regard the tables as the same.

Case 2. Three families are partitioned in (4, 1) fashion and one is partitioned in (3, 2) fashion.
There are 4 ways to choose which family to be in (3, 2) fashion. The 3-group must sit with one of the three parents from another family, while the 2-group must sit with the remaining two parents. There are 2 ways to choose a parent of each family, and 3 ways to choose a child of the family in (3, 2) fashion to be put in the 2-group. So the number of ways is $4 \times 3 \times 2^4 \times 3 = 576$.

Case 3. Two families are partitioned in (4, 1) fashion and two are partitioned in (3, 2) fashion.
There are $\binom{4}{2} = 6$ ways to choose which two families to be in (3, 2) fashion. The two 2-groups must sit together, while the 3-groups must sit with one of the two parents from the other families. There are 2 ways to choose a parent of each family, and 3 ways to choose a child of each family in (3, 2) fashion to be put in the 2-groups. So the number of ways is $6 \times 2 \times 2^4 \times 3^2 = 1728$.

It is easy to check that no other cases are possible. Since the tables are different, there are $5! = 120$ ways to arrange the tables. Therefore the answer is
$$120 \times (16 + 576 + 1728) = 278400.$$