>>> p = "one two three four five six seven three four eight"
 >>> s = set()
 >>> print ' '.join(w for w in p.split() if not (w in s or s.add(w)))
one two three four five six seven eight

which is absolutely horrible because it mutates the set within
the list comprehension.  The passable solution would use a
for-loop to iterate over each word in the paragraph, emitting it
if it hadn't already been seen.  Maintain those words in set, so
your words know how not to be seen. ("Mr. Nesbitt, would you
please stand up?")

This also assumes your paragraph consists only of words and
whitespace.  But since you posted your previous homework-sounding
question on stripping out non-word/whitespace characters, you'll
want to look into using a regexp like "[\w\s]" to clean up the
cruft in the paragraph.  Neither solution above preserves non
white-space/word characters, for which I'd recommend using a
re.sub() with a callback.  Such a callback class might look
something like

 >>> class Dedupe:
...     def __init__(self):
...             self.s = set()
...     def __call__(self, m):
...             w = m.group(0)
...             if w in self.s: return ''
...             self.s.add(w)
...             return w
...
 >>> r.sub(Dedupe(), p)

where I leave the definition of "r" to the student.  Also beware
of case-differences for which you might have to normalize.

You'll also want to use more descriptive variable names than my
one-letter tokens.

 >>> p = re.compile(r'(.+) .*\1')
 >>> s = p.search("python one two one two python")
 >>> s.groups()
('python',)
 >>> s = p.search("python one two one two python one")
 >>> s.groups() # guess what happened to the 2nd "one"...
('python one',)

and even once you have the list of theoretical duplicates (by
changing the regexp to r'\b(\w+)\b.*?\1' perhaps), you still have
to worry about emitting the first instance but not subsequent
instances.