import inspect

def foo(a, b, x='blah'):
    pass

print(inspect.signature(foo))
# (a, b, x='blah')

Python 3.5+ recommends inspect.signature().

In general the signature of a function is defined by the number and type of input arguments the function takes and the type of the result the function returns.

As an example in C++ consider the following function:

int multiply(int x, int y){
    return x*y;
}

The signature of that function, described in an abstract way would be the set {int, int, int}.

As Python is a weakly typed language in general the signature is only given by the amount of parameters. But in newer Python versions type hints were introduced which clarify the signature:

def multiply(x: int, y: int) -> int:
    return x*y

Python3, 302 bytes

from itertools import*
V=lambda v:not any(not v[i][0] and v[i-1][0]for i in range(len(v))if i)
def f(a,b,c):
 P=product(*[*[[(None,t)]+[[]]*k for n,t,k in a],*[[(None,t),(n,t)]+[[]]*k for n,t,k in b],*[[(n,t)]+[[]]*k for n,t,k in c]])
 return[*map(eval,{str(K)for i in P if V(K:=[j for j in i if j])})]

Try it online!

For those interested, here is a larger solution which parses the function signature:

import ast
from itertools import*
A=lambda x:x.annotation.id
H=lambda v:not any(not v[i][0] and v[i-1][0]for i in range(len(v))if i)
def f(s):
 a=ast.parse(s).body[0].args
 V=['kw_defaults','defaults']
 d=sum(len(getattr(a,i))for i in V)
 l=sum(len(i)for j in a._fields if(i:=getattr(a,j))and j not in V)
 K=0
 P=product(*[*[[(None,A(i))]for i in a.posonlyargs if(K:=K+1)],*[[(None,A(i)),(i.arg,A(i))]+[[]]*(((K:=K+1))>l-d)for i in a.args],*[[(i.arg,A(i)),[]]for i in a.kwonlyargs]])
 return[K for i in P if H(K:=[j for j in i if j])]

Try it online!