= Solution
\b[Yes: a deterministic LOCC protocol exists for every allowed value of the parameter.] Here is an explicit implementation, so existence does not depend on merely quoting a pure-state conversion theorem.
First convert the input <Bell state> to $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$ by <local unitary operations>:
$$
(Z_A\otimes X_B)|\Psi^-\rangle=|\Phi^+\rangle.
$$
Write $c=\cos\phi$ and $s=\sin\phi$. Alice performs the two-outcome <quantum measurement> with <Kraus operators>
$$
M_0=\begin{pmatrix}c&0\\0&s\end{pmatrix},\qquad
M_1=\begin{pmatrix}s&0\\0&c\end{pmatrix}.
$$
They define a physical <quantum measurement> because
$$
M_0^\dagger M_0+M_1^\dagger M_1=(c^2+s^2)I=I.
$$
Acting on $|\Phi^+\rangle$, the unnormalized branch vectors are
$$
(M_0\otimes I)|\Phi^+\rangle=\frac{c|00\rangle+s|11\rangle}{\sqrt2},\qquad
(M_1\otimes I)|\Phi^+\rangle=\frac{s|00\rangle+c|11\rangle}{\sqrt2}.
$$
Both branches have <probability> $1/2$. Alice communicates the outcome to Bob. If it is one, both apply the <Pauli X gate>; this turns the normalized second branch into $c|00\rangle+s|11\rangle$. If it is zero, neither applies this correction. Finally Bob applies a <Pauli X gate> in either branch, obtaining
$$
\boxed{c|01\rangle+s|10\rangle.}
$$
Every outcome produces the same target <pure state>, so this is deterministic <local operations and classical communication>, not a successful branch selected by <postselection>. This <two-outcome LOCC dilution of a Bell pair> remains valid at $\phi=0$, where the output is a <product state>, and at $\phi=\pi/4$, where it is another <Bell state>.
The target <Schmidt coefficients> are $c,s$, while those of the input are both $1/\sqrt2$. For an interior value $0<\phi<\pi/4$, <local unitary operations> alone could not change these <Schmidt coefficients>; the local <quantum measurement> is the step that reduces the <entanglement>.
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