= Solution
Use base-two <logarithms>, so <entanglement entropy> is measured in bits. Write $h_2(p)=-p\log_2p-(1-p)\log_2(1-p)$ for the <binary entropy>, with $0\log_20=0$. Taking a <partial trace> over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party <reduced density matrices> of the first <pure state> are
$$
\rho_A=\rho_B=\rho_C=p|0\rangle\langle0|+(1-p)|1\rangle\langle1|.
$$
Each division into two groups is one party versus the other two. The <Schmidt coefficients> are $\sqrt p$ and $\sqrt{1-p}$, so \b[the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$]. For the second <pure state>, set $p=1/2$: every one-party <reduced density matrix> is $I/2$, and \b[all three entanglement entropies are one bit].
For every pair $XY$, the first <pure state> instead gives the <reduced density matrix>
$$
\rho_{XY}=p|00\rangle\langle00|+(1-p)|11\rangle\langle11|.
$$
This is a <separable quantum state>, explicitly a <convex combination> of two <product states>. Its <relative entropy of entanglement> is therefore zero: in $E_R(\rho)=\inf_{\sigma\in\mathrm{Sep}}D(\rho\|\sigma)$, choose $\sigma=\rho$ to attain zero, while <nonnegativity of quantum relative entropy> excludes a negative value. The same argument with equal weights applies to the second <pure state>. Thus
$$
\boxed{E_R(\rho_{AB})=E_R(\rho_{AC})=E_R(\rho_{BC})=0\quad\text{for both states}.}
$$
The pairwise <density matrices> contain classical correlations, despite the <entanglement> of the global <pure state>; discarding a party loses the coherence between its two branches.
The <entanglement entropy> of $n$ independent source copies across any cut is $nh_2(p)$, whereas $f(n)$ copies of the target <GHZ state> have <entanglement entropy> $f(n)$. <Average monotonicity of pure-state entanglement entropy> under <LOCC> gives $P(n)f(n)\leq nh_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n'/n\to1$. For asymptotically faithful outputs, a <continuity bound for quantum conditional entropy> with trivial conditioning changes these inequalities only by $o(n)$. Hence \b[the reversible leading-order rate is]
$$
\boxed{f(n)=nh_2(p)+o(n),\qquad \lim_{n\to\infty}\frac{f(n)}n=h_2(p).}
$$
The zero pairwise <relative entropies of entanglement> are consistent with this conversion: no independent pairwise resource has to be supplied or removed.
There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the <Schmidt rank> of the exact state of $n'$ source copies is $2^{n'}$, whereas $f$ target <GHZ states> have <Schmidt rank> $2^f$. Every refined <LOCC> branch acts by a product of local <linear operators> and cannot increase this <Schmidt rank>. For $p\ne1/2$, the displayed rate has $f<n'$ for sufficiently large blocks when $n'/n\to1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The <exact dilution obstruction from Schmidt rank> is avoided in the usual asymptotic formulation by allowing outputs approaching the desired <pure state>. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known <product state>, the rate is zero, and local preparation supplies any desired number of source copies.
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