= Solution
Write $a=|\alpha|^2$, $b=|\beta|^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party gives
$$
|\Psi\rangle=\sum_{x\in\{0,1\}^4}\alpha^{z(x)}\beta^{4-z(x)}|x\rangle_A|x\rangle_B,
$$
where $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate <eigenspace>, rather than measuring her four bits separately. Her <projective measurement> has projectors $\Pi_k=\sum_{z(x)=k}|x\rangle\langle x|$, and the <Born rule> gives
$$
\boxed{p_k=\binom4k a^k b^{4-k},\qquad k=0,1,2,3,4.}
$$
Alice sends $k$ to Bob. Up to the common phase of $\alpha^k\beta^{4-k}$, their conditional <pure state> is
$$
|T_k\rangle=\frac1{\sqrt{d_k}}\sum_{z(x)=k}|x\rangle_A|x\rangle_B,\qquad d_k=\binom4k.
$$
This is <Schmidt projection entanglement concentration>: its equal <Schmidt coefficients> give a <maximally entangled state> of <Schmidt rank> $d_k$.
For $k=0$ or $4$, $d_k=1$, and the branch is a <product state>. No <Bell pair> can be extracted using <local operations and classical communication>. The respective <probabilities> are $b^4$ and $a^4$.
For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a <local unitary operation> extending the basis relabelling
$$
|x_j\rangle\longmapsto |j_1j_0\rangle\otimes|00\rangle,\qquad j=2j_1+j_0.
$$
Such a unitary exists because these are four distinct <orthonormal> basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leaves
$$
\frac12\sum_{j_1,j_0=0}^1|j_1j_0\rangle_A|j_1j_0\rangle_B
=|\Phi^+\rangle_{A_1B_1}\otimes|\Phi^+\rangle_{A_0B_0}.
$$
Thus these branches give exactly two <Bell pairs>, with <probabilities> $4ab^3$ and $4a^3b$, and no further <quantum measurement> is needed.
For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local <POVM> gives two <Bell pairs> with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the <Kraus operators>
$$
K_r=\frac1{\sqrt2}\sum_{j\notin T_r}|x_j\rangle\langle x_j|,\qquad r=0,1,2.
$$
On the six-dimensional support, each label is retained by two operators, so $\sum_{r=0}^2K_r^\dagger K_r=\Pi_2$. Add $K_\perp=I-\Pi_2$ to define a complete <quantum measurement> on Alice's four qubits; that extra outcome has <probability> zero on this branch. Each nonzero outcome has conditional <probability> $(1/2)(4/6)=1/3$, and gives the normalized state
$$
\frac12\sum_{j\notin T_r}|x_j\rangle_A|x_j\rangle_B.
$$
Alice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two <Bell pairs>. The joint <probability> of $(k,r)=(2,r)$ is $p_2/3=2a^2b^2$ for each $r$.
For an explicit local implementation of this <POVM>, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,
$$
|x_j\rangle\longmapsto |x_j\rangle\otimes\frac1{\sqrt2}\sum_{r:\,j\notin T_r}|r\rangle.
$$
It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a <deterministic reduction of maximally entangled Schmidt rank> from six to four.
The complete output distribution for this strategy is consequently
$$
\boxed{\Pr(2\text{ Bell pairs})=1-a^4-b^4,\qquad \Pr(0\text{ Bell pairs})=a^4+b^4.}
$$
The mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has <Schmidt rank> at most six, whereas three independent <Bell pairs> require <Schmidt rank> eight. In any refined <LOCC> branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its <rank>.
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