您可以使用str.extract：

df[['Addressess', 'Post Code']] = df['Addressess'].str.extract(r'(.*?)\s*(\bEH\d+[\s\w]*)?\W*$')

regex demo
或str.split，如果至少有一行具有邮政编码：

df[['Addressess', 'Post Code']] = df['Addressess'].str.split(r'\s*(?=\bEH\d*)', n=1, expand=True)

regex demo
输出：

Addressess Post Code
0       10 Pentland Drive, Comiston, Edinburgh,  EH10 6PX
1                       Moray Place, Edinburgh,       EH3
2                     Carlton Street, Edinburgh       NaN
3  The Bourse Apartments, 47 Timber Bush, Leith   EH6 6QH

使用简单的正则表达式匹配：

df['Post Code'] = df['Addressess'].str.extract('(EH.+)').fillna('')

Addressess  Post Code
0  10 Pentland Drive, Comiston, Edinburgh, EH10 6PX.  EH10 6PX.
1                        Moray Place, Edinburgh, EH3        EH3
2                          Carlton Street, Edinburgh           
3  The Bourse Apartments, 47 Timber Bush, Leith E...    EH6 6QH

下面是使用正则表达式正前瞻的方法：

df['Addressess'].str.split(r'[0-9A-Za-z,] (?=EH)',expand=True).rename({0:'Addressess',1:'Post Code'},axis=1)