= Solution
Let $D$ denote Alice's unknown input qubit and $A$ her GHZ qubit. Expanding the four-qubit state in the <Bell state>[Bell basis] of $D,A$ gives
$$
|\alpha\rangle_D|\operatorname{GHZ}\rangle_{ABC}
=\frac12\Big[
|\Phi^+\rangle_{DA}(a|00\rangle+b|11\rangle)_{BC}
+|\Phi^-\rangle_{DA}(a|00\rangle-b|11\rangle)_{BC}
$$
$$
\qquad
+|\Psi^+\rangle_{DA}(b|00\rangle+a|11\rangle)_{BC}
+|\Psi^-\rangle_{DA}(b|00\rangle-a|11\rangle)_{BC}
\Big],
$$
where a global sign in the last term is immaterial.
For outcomes $\Phi^+$ and $\Phi^-$, Bob and Charlie already have
$$
|\phi_1\rangle=a|00\rangle+b|11\rangle,
\qquad
|\phi_2\rangle=a|00\rangle-b|11\rangle.
$$
For either $\Psi$ outcome, both apply the <Pauli X gate>. Since $X\otimes X$ exchanges $|00\rangle$ and $|11\rangle$, the $\Psi^+$ branch becomes $|\phi_1\rangle$ and the $\Psi^-$ branch becomes $|\phi_2\rangle$, up to a global phase. This proves the <Bell measurement on one qubit and one leg of a GHZ state> result.
Finally, tracing out Charlie gives
$$
\rho_B=|a|^2|0\rangle\langle0|+|b|^2|1\rangle\langle1|
$$
for either sign. This reduced state has rank two exactly when $a\ne0$ and $b\ne0$. By the <entanglement criterion for a two-term correlated state>,
$$
\boxed{|\phi_1\rangle\text{ and }|\phi_2\rangle
\text{ are entangled iff }a\ne0\text{ and }b\ne0.}
$$
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