The sample code presented below produces error (screenshot available in 
attachment) on Windows 7 RC x64 with latest version of Python 2.6 
installed. Both 32bit and 64bit versions of Python 2.6 are affected by 
this issue. Python 3 executes this code flawlessly.
The problem seems to be related to stdout redirection. If the "stdout = 
PIPE" is omitted in the first statement of main() function, code 
executes properly but we don't have access to the processes output. If 
the line "Popen(["dir"], stdout = PIPE, shell = True)" is changed to 
"sin, sout = os.popen2(["dir"])" and os module is imported everything 
works fine. It doesn't matter what command is executed ("dir", "bcedit", 
"mspaint" etc)


import sys
from subprocess import Popen, PIPE
from multiprocessing import Process

def myProcess():
    Popen(["dir"], shell = True) # ERROR :(

def main():
    Popen(["dir"], stdout = PIPE, shell = True)
    Process(target = myProcess).start()
    return 0

if __name__ == "__main__":
    sys.exit(main())

I cannot reproduce it on Win7 32bit (OS Name: Microsoft Windows 7
Enterprise; OS Version: 6.1.7600 N/A Build 7600) with Python 2.6.3rc1.
Attached there's the file I used (a slightly different version of what
you posted, your version worked too), this is the output I got:

cmd>python issue6773.py
"hello from myProcess"

cmd>

I cannot reproduce it on Windows 7 x64 Ultimate either.
sys.getwindowsversion() -> (6, 1, 7600, 2, '')

Probably this bug was in Win7 RC and not in python. Thanks for your time. 
Closing.