>>> from collections import OrderedDict
>>> od = OrderedDict([('a', 1), ('b', 2)])
>>> od.move_to_end('a')
>>> od
OrderedDict([('b', 2), ('a', 1)])
>>> dict(od)
{'a': 1, 'b': 2}

This affects also PEP-468.

>>> def f(**kwargs): return kwargs
... 
>>> f(**od)
{'a': 1, 'b': 2}

And PEP-520.

>>> type('A', (), od).__dict__
mappingproxy({'a': 1, 'b': 2, '__module__': '__main__', '__dict__': <attribute '__dict__' of 'A' objects>, '__weakref__': <attribute '__weakref__' of 'A' objects>, '__doc__': None})

FWIW, the PEP-520 example isn't exactly right. PEP-520 is about the class definition namespace, not the namespace object passed to type(). That always does the right thing, since the default class definition namespace is dict (which happens to be ordered in 3.6+).

That said, a test that explicitly calls type() with an OrderedDict, as you have shown, is still a good idea since type() always changes the namespace to a dict.

Perhaps another good test of the same thing would be with a metaclass that returns an OrderedDict from __prepare__():

  class Meta(type):
      @classmethod
      def __prepare__(meta, name, bases, **kwargs):
          return OrderedDict()

  class Spam(metaclass=Meta):
      a = 1
      b = 2
      locals().move_to_end('a')

Spam.__dict__

Just keep in mind that neither of those is specific to PEP-520.

Perhaps another good test of the same thing would be with a metaclass that returns an OrderedDict from __prepare__():

It is not relating to this issue: there are no conversion happened.
Should I add the test in this PR?

There is a conversion. See builtin___build_class__ in Python/bltinmodule.c (and the LOAD_BUILD_CLASS target in Python/ceval.c). Specifically, the metaclass (e.g. the builtin type) is instantiated using the namespace from the class definition. The metaclass copies that namespace into a new dict. So the following two bits of code are equivalent:

# using a class definition

  ns = OrderedDict()

  class Meta(type):
      def __prepare__(*args, **kwargs):
          return ns

  class Spam(metaclass=Meta):
      a = 1
      b = 2
      ns.move_to_end('a')

# using the metaclass directly

  ns = OrderedDict()
  ns['a'] = 1
  ns['b'] = 2
  ns.move_to_end('a')
  Spam = Meta('Spam', (object,), ns)

In both cases Spam.__dict__ will be a proxy for a new dict copied from ns.

Conversion is happens when type() (or metaclass()) is called.
And I added test for it already.
https://github.com/python/cpython/pull/8624/files#diff-7ba610fbe5686a1d67c5504312be4817R1977

Why regression test for this bugfix need to use __prepare__?

Conversion is happens when type() (or metaclass()) is called.

Right. So an extra test to cover the __prepare__ case is somewhat redundant. I only suggested it because type() is called implicitly and the test would make the conversion case clear. However, it was only a suggestion for you to consider, not something I considered necessary. :) I'm fine with not adding a test that uses __prepare__.

New changeset 2aaf98c by Miss Islington (bot) (INADA Naoki) in branch 'master':
bpo-34320: Fix dict(o) didn't copy order of dict subclass (GH-8624)
2aaf98c

New changeset 12e3e80 by Miss Islington (bot) in branch '3.7':
bpo-34320: Fix dict(o) didn't copy order of dict subclass (GH-8624)
12e3e80

New changeset d45a961 by Miss Islington (bot) in branch '3.6':
[3.6] bpo-34320: Fix dict(o) didn't copy order of dict subclass (GH-8624) (GH-9583)
d45a961