Find anagrams of a word in a list
The most pythonic way (IMO) to check if two words are anagrams of each other is to check if the number of times each letter appears in each words is the same. For this you can use collections.Counter:

from collections import Counter

&gt;&gt;&gt; Counter("test")
Counter({'e': 1, 's': 1, 't': 2})


With this we can re-write your function to :

def anagrams(word, words):
    result = []
    counter_word = Counter(word)
    for other_word in words:
        if Counter(other_word) == counter_word:
            result.append(other_word)
    return result


This is a lot more readable. in addition, here are some more comments on your code itself:


source = copy.deepcopy(list(i)) is not needed. Since strings are immutable, just doing list(i) would have been fine.
In if(j in source): there are two things:


Python has an official style-guide, PEP8, which is well worth a read. It recommends not using parenthesis in if (and while) statements, so this should be if j in source
in is \$\mathcal{O}(n)\$ for lists and strings, so this is actually quite an expensive way to do it. The better way is using a dictionary (collections.Counter is a dictionary plus some nice stuff).

In for i in wordList, again, PEP8 has something to say, it should be word_list, or even easier words. Also i is not a good name.


Depending on what you want to do afterwards with the anagrams, you could also make this a generator (which you can iterate over and which continuesly yields anagrams):

def anagrams(word, words):
    counter_word = Counter(word)
    for other_word in words:
        if Counter(other_word) == counter_word:
            yield other_word