Well, ¬d ∨ d is true, and second, we have theabsorption law ¬c ∧ d ∨ c = d ∨ c, so that we get

¬φ =
¬a ∧ b ∧ c ∧ d ∨
a ∧ (¬b ∧ c ∧ d ∨
b ∧ (¬c ∧ d ∨
c ∧ (¬d ∨ d )))
=
¬a ∧ b ∧ c ∧ d ∨
a ∧ (¬b ∧ c ∧ d ∨
b ∧ (¬c ∧ d ∨
c ))
=
¬a ∧ b ∧ c ∧ d ∨
a ∧ (¬b ∧ c ∧ d ∨
b ∧ (c ∨ d))

However, the simplifications were not really required, and they also do not matter for the construction of the BDD in the next part of the exam problem (since BDDs are the same for all equivalent formulas).