 Solution

An easy lambda term must be equatable consistently with every closed lambda term. Choose the proposed equation $I=K$. Applying it to $I$ gives $I=\lambda y.I$. Applying that equation to an arbitrary closed $V$ gives $V=I$. In particular the two Church Booleans become equal; then applying them to arbitrary $A,B$ gives $A=B$. The extended equational theory is trivial, hence inconsistent in the required sense. Therefore $I$ is not easy.