= Solution
Let Alice's two bits be $(x,z)$ and Bob's bit be $b$. Alice applies
$$
X_A^xZ_A^z,
$$
while Bob applies
$$
X_B^b.
$$
They then send their qubits to Charlie. Acting on
$$
|\chi_1^+\rangle
=\frac{|000\rangle+|111\rangle}{\sqrt2},
$$
these operations produce, up to an irrelevant global phase,
$$
\frac{|x,b,0\rangle+(-1)^z
|1\mathbin\oplus x,1\mathbin\oplus b,1\rangle}{\sqrt2}.
$$
The complete encoding table is
$$
\begin{array}{c|c|c}
(x,z)&b&\text{state measured by Charlie}\\ \hline
(0,0)&0&|\chi_1^+\rangle\\
(0,1)&0&|\chi_1^-\rangle\\
(0,0)&1&|\chi_2^+\rangle\\
(0,1)&1&|\chi_2^-\rangle\\
(1,0)&0&|\chi_4^+\rangle\\
(1,1)&0&|\chi_4^-\rangle\\
(1,0)&1&|\chi_3^+\rangle\\
(1,1)&1&|\chi_3^-\rangle
\end{array}
$$
where a possible overall minus sign in the two $x=z=1$ cases has no observable effect.
Charlie performs a projective measurement in the eight-state GHZ basis. The sign $+$ or $-$ reveals $z$. The basis index reveals $(x,b)$ according to
$$
1\leftrightarrow(0,0),
\quad
2\leftrightarrow(0,1),
\quad
3\leftrightarrow(1,1),
\quad
4\leftrightarrow(1,0).
$$
He therefore reconstructs Alice's two bits and Bob's one bit with certainty, which is the <three-party dense coding with a GHZ state> protocol.
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