What causes [*a] to overallocate?

[*a] is internally doing the C equivalent of:

1. Make a new, empty list
2. Call newlist.extend(a)
3. Returns list.

So if you expand your test to:

from sys import getsizeof

for n in range(13):
    a = [None] * n
    l = []
    l.extend(a)
    print(n, getsizeof(list(a)),
             getsizeof([x for x in a]),
             getsizeof([*a]),
             getsizeof(l))

Try it online!

you’ll see the results for getsizeof([*a]) and l = []; l.extend(a); getsizeof(l) are the same.

This is usually the right thing to do; when extending you’re usually expecting to add more later, and similarly for generalized unpacking, it’s assumed that multiple things will be added one after the other. [*a] is not the normal case; Python assumes there are multiple items or iterables being added to the list ([*a, b, c, *d]), so overallocation saves work in the common case.

By contrast, a list constructed from a single, presized iterable (with list()) may not grow or shrink during use, and overallocating is premature until proven otherwise; Python recently fixed a bug that made the constructor overallocate even for inputs with known size.

As for list comprehensions, they’re effectively equivalent to repeated appends, so you’re seeing the final result of the normal overallocation growth pattern when adding an element at a time.

To be clear, none of this is a language guarantee. It’s just how CPython implements it. The Python language spec is generally unconcerned with specific growth patterns in list (aside from guaranteeing amortized O(1) appends and pops from the end). As noted in the comments, the specific implementation changes again in 3.9; while it won’t affect [*a], it could affect other cases where what used to be “build a temporary tuple of individual items and then extend with the tuple” now becomes multiple applications of LIST_APPEND, which can change when the overallocation occurs and what numbers go into the calculation.