将join与\b\b一起用于单词边界，将|用于正则表达式or：

lis=['Engineer','Executive','Teacher']

pat = '|'.join(r"\b{}\b".format(x) for x in lis)

df['Welfare_Cost'] = np.where(((df['Education']=='Masters') & 
                              (df['Occupation'].str.contains(pat))),
                              100,50)

或者：

df['Welfare_Cost'] = np.where(((df['Education']=='Masters') & 
                              (df['Occupation'].str.contains('|'.join(lis)))),
                              100,50)

print (df)
             Occupation  Education  Welfare_Cost
0         Engineer High     School            50
1          Neurosurgeon    Masters            50
2   Electrical Engineer    Masters           100
3   Mechanical Engineer    Masters           100
4     Software Engineer    Masters           100
5              Engineer    Masters           100
6    Business Executive    Masters           100
7       Sales Executive  Bachelors            50
8          Neurosurgeon    Masters            50
9   Electrical Engineer        NaN            50
10           Accountant  Bachelors            50
11      Sales Executive    Masters           100

可能存在差异，参见变更数据-\b\b匹配字符串，无子字符串：

lis=['Engineer','Executive','Teacher']

df['Welfare_Cost1'] = np.where(((df['Education']=='Masters') & 
                              (df['Occupation'].str.contains('|'.join(lis)))),
                              100,50)

pat = '|'.join(r"\b{}\b".format(x) for x in lis)

df['Welfare_Cost2'] = np.where(((df['Education']=='Masters') & 
                              (df['Occupation'].str.contains(pat))),
                              100,50)

print (df)
              Occupation  Education  Welfare_Cost1  Welfare_Cost2
0          Engineer High     School             50             50
1           Neurosurgeon    Masters             50             50
2   Electrical Engineers    Masters            100             50
3    Mechanical Engineer    Masters            100            100
4      Software Engineer    Masters            100            100
5               Engineer    Masters            100            100
6     Business Executive    Masters            100            100
7        Sales Executive  Bachelors             50             50
8           Neurosurgeon    Masters             50             50
9    Electrical Engineer        NaN             50             50
10            Accountant  Bachelors             50             50
11      Sales Executives    Masters            100             50

在您的示例中，过滤器列表只包含职业名称的基本部分（语义上），因此检查str.endswith就足够了：

df['Welfare_Cost']=np.where((df['Education']=='Masters') & df['Occupation'].str.endswith(tuple(lis)),100,50)

Occupation    Education  Welfare_Cost
0              Engineer  High School            50
1          Neurosurgeon      Masters            50
2   Electrical Engineer      Masters           100
3   Mechanical Engineer      Masters           100
4     Software Engineer      Masters           100
5              Engineer      Masters           100
6    Business Executive      Masters           100
7       Sales Executive    Bachelors            50
8          Neurosurgeon      Masters            50
9   Electrical Engineer         None            50
10           Accountant    Bachelors            50
11      Sales Executive      Masters           100