The boolean expression from a circuit
Using De Morgan's theorem: (A*B)' = A' + B'

The output of AND3 gate in the image should be (a'+b)*(b*(c'+d')) instead of (a'+b)*(b*(c'*d'))

Which is how I assume you arrived at the edit part of your question. Now expanding the above expression, you get:

a'bc' + a'bd' + bc' + bd'

The above expression can be further simplified by grouping the terms:

bc'(a'+1) + bd'(a'+1)

And further using the Annulment law of Boolean Algebra, you get:

b (c' + d') as your final simplified expression.

Please let me know if this answered your question.