if and else in python lambda expression

First, let's remove this line as it doesn't do anything:

lambda x: 'big' if x > 100 else 'small'

This lambda expression is defined but never called. The fact that it's argument is also called x has nothing to do with the rest of the code.

Let's look at what remains:

a = 4
b = 7

x = lambda: a if 1 else b

print(x())

Here x becomes a function as it contains code. The lambda form can only contain expressions, not statements, so it has to use the expression form of if which is backward looking:

true-result if condition else false-result

In this case the condition is 1, which is always true, so the result of the function x() is always the value of a, assigned to 4 earlier in the code. Effectively, x() acts like:

def x():
    return a

Understanding the differences between expressions and statements is key to understanding code like this.

Your x is always equals to 4, as it takes no arguments and if 1 is always True. Then you have lambda expression that's not assigned to any variable, neither used elsewhere. Eventualy, you print out x, which is always 4 as I said above.

P.S. I strongly suggest you to read Using lambda Functions from Dive into Python

Let me translate that for you.

You assign to x a lambda function with no arguments. Because 1 always evaluates as true, you always return the externally defined variable a, which evaluates as 4. Then, you create a lambda function with one argument x, which you don't assign to a variable/access name, so it is lost forever. Then, you call function x, which always returns a. Output is 4.