 Solution

Yes: deterministic entanglement swapping gives the required singlet. Denote Bob's two local qubits by $B_1,B_2$. Known local unitary operations first convert the two given spin singlet states into $|{\Phi }^{+}{⟩}_{A{B}_1}$ and $|{\Phi }^{+}{⟩}_{B_{2}C}$. For example applying $ZX$ to the second qubit of a singlet gives $|{\Phi }^{+}⟩$, where $X,Z$ are Pauli operators.

Use the binary-labelled Bell statesThey are ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ for labels $(0,0),(1,0),(0,1),(1,1)$. Reordering registers to group Bob's qubits, direct expansion givesAll four basis vectors are real in this convention; otherwise the paired expansion must account for complex conjugation. Bob performs the Bell-basis measurement on $B_1{B}_2$. This joint operation is local to him, so it is allowed by LOCC. Each outcome occurs with probability $1/4$ and leaves $AC$ in the corresponding Bell state.

Bob sends $(p,q)$ to Charlie. Charlie applies $Z^p{X}^{-q}$ to obtain $|{\Phi }^{+}{⟩}_{AC}$, and then $XZ$ to obtainThe correction works for every outcome, so no postselection or successful-outcome assumption is needed. This is entanglement swapping, equivalently teleportation of $B_1$ to $C$ while preserving its entanglement with $A$. If Bob's measurement result is ignored, the average $AC$ state is $I_4/4$; the protocol therefore does not supply entanglement usable without the classical outcome or violate the no-communication theorem.