Taking advantage of numpy broadcasting, you can do it easily:

(df[['d']].to_numpy() == df.to_numpy())[:, :-1].any(axis=1)

Output:

array([False,  True, False,  True])

Using numpy is about as fast as you can get.

Try:

out = (df[['a','b','c']].T==df['d']).any()

Output:

0    False
1     True
2    False
3     True
dtype: bool

Use eq to broadcast equality comparison across the columns. Then check if there are any matches in the row:

df.drop(columns='d').eq(df['d'], axis=0).any(axis=1).array
<PandasArray>
[False, True, False, True]
Length: 4, dtype: bool

Speedwise, a numpy option is best.

Try with

df.eq(df.pop('d'),axis=0).any(1)#.values
0    False
1     True
2    False
3     True
dtype: bool

Need to add 'values':

df.apply(lambda x: x['d'] in x[['a','b','c']].values, axis=1)
Out[11]: 
0    False
1     True
2    False
3     True
dtype: bool

What you've typed is checking whether x['d'] is in the set of keys of x[['a','b','c']] not values. Dictionaries work in similar ways:

1 in {'a': 1, 'b': 2}
Out[30]: False
1 in {'a': 1, 'b': 2}.values()
Out[31]: True