= Solution
Here is a <tripartite Schmidt projection entanglement concentration> protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local <qubits> of each party gives
$$
|\Psi_1\rangle^{\otimes n}=\sum_{x\in\{0,1\}^n}p^{(n-|x|)/2}q^{|x|/2}|x\rangle_A|x\rangle_B|x\rangle_C,
$$
where $|x|$ counts ones. Alice performs a <projective measurement> of that count, without learning the individual string. The <orthogonal projector> for outcome $k$ is $\Pi_k=\sum_{|x|=k}|x\rangle\langle x|$. By the <Born rule>, its <probability> and conditional <pure state> are
$$
P_k=\binom nk p^{n-k}q^k,\qquad
|T_k\rangle=\frac1{\sqrt{d_k}}\sum_{|x|=k}|x\rangle_A|x\rangle_B|x\rangle_C,\qquad d_k=\binom nk.
$$
Alice sends $k$ by <classical communication>; all three parties use <local unitary operations> to relabel the same ordered type class as $0,\ldots,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual <qubits> instead would have destroyed precisely this coherence.
To obtain an integer number $f$ of target <GHZ states>, choose $m=2^f\leq d_k$. The following exact <LOCC> step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq\{0,\ldots,d_k-1\}$, Alice uses the <Kraus operator>
$$
K_S=\frac{\Pi_S}{\sqrt{\binom{d_k-1}{m-1}}},\qquad \Pi_S=\sum_{r\in S}|r\rangle\langle r|.
$$
Every label occurs in $\binom{d_k-1}{m-1}$ such subsets, proving $\sum_SK_S^\dagger K_S=I$ on the occupied subspace. Thus this is a complete <quantum measurement>. Each outcome has <probability> $m/[d_k\binom{d_k-1}{m-1}]=1/\binom{d_k}{m}$, and leaves the normalized <pure state> $m^{-1/2}\sum_{r\in S}|r,r,r\rangle$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into \b[exactly $f$ independent copies of the target GHZ state]. Local extensions of these operations handle unused basis vectors and do not affect the protocol.
For example, taking $f(k)=\lfloor\log_2d_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a <binomial distribution> with parameters $(n,q)$, the <weak law of large numbers> gives $K/n\to q$ in probability. <Stirling's formula>, uniformly when $k/n$ stays away from zero and one, gives
$$
\log_2\binom nk=nh_2(k/n)-\frac12\log_2\bigl(2\pi n(k/n)(1-k/n)\bigr)+O(n^{-1}).
$$
By continuity of the <binary entropy>, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.
A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, take
$$
f_n=\left\lfloor nh_2(p)-n^{3/4}\right\rfloor.
$$
On $|K-nq|\leq n^{2/3}$, the bounded derivative of the <binary entropy> in a neighborhood of the fixed $q\in(0,1)$ and <Stirling's formula> give $\log_2d_K\geq nh_2(q)-C_qn^{2/3}-O(\log n)\geq f_n$. <Chebyshev's inequality> bounds the complementary <probability> by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset <quantum measurement> with $m=2^{f_n}$. Thus
$$
\boxed{P(\text{success})\longrightarrow1,\qquad f_n/n\longrightarrow h_2(p).}
$$
This matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.
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