= Solution
\b[Yes, the conversion can be deterministic using <local operations and classical communication>.] Put $s=\sin\theta$ and $c=\cos\theta$. The initial state's squared <Schmidt coefficients> are $(1/2,1/2)$, while the target's decreasingly ordered squared <Schmidt coefficients> are $(c^2,s^2)$. Since $0<\theta<\pi/4$, one has $1/2<c^2$ and both pairs sum to one. Thus
$$
(1/2,1/2)\prec(c^2,s^2).
$$
The direction of this <majorization> is the allowed one in <Nielsen's pure-state conversion theorem>: maximal entanglement can be reduced by deterministic <LOCC>.
Here is an explicit protocol, so no conversion theorem need be assumed. The sender first applies a <Pauli Z gate> to remove the initial Bell-state minus sign. Starting from $|\Phi^+\rangle$, she performs a two-outcome local measurement with <Kraus operators>
$$
M_0=\begin{pmatrix}s&0\\0&c\end{pmatrix},
\qquad
M_1=\begin{pmatrix}c&0\\0&s\end{pmatrix}.
$$
They define a valid measurement because $M_0^\dagger M_0+M_1^\dagger M_1=I$. The two unnormalized states after measurement are
$$
(M_0\otimes I)|\Phi^+\rangle=\frac{s|00\rangle+c|11\rangle}{\sqrt2},
\qquad
(M_1\otimes I)|\Phi^+\rangle=\frac{c|00\rangle+s|11\rangle}{\sqrt2}.
$$
Each outcome has probability $1/2$. For outcome zero, the normalized state is already $s|00\rangle+c|11\rangle$. For outcome one, both parties apply <Pauli X gates>, obtaining the same state. The sender communicates only which outcome occurred. Finally the receiver applies a <Pauli X gate>, giving $s|01\rangle+c|10\rangle$ in either branch. This is <deterministic two-qubit entanglement dilution>, with total success probability one; it does not rely on postselection.
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