Step 1: Ideal circuit and target state
We start in \(|0000\rangle\). The circuit (right‐to‐left) is
\[
H_2,\;
\CNOT_{2\to1},\;
H_0,\;
\CNOT_{0\to3}\,.
\]
- \(H_2\) on qubit 2 creates \(\tfrac1{\sqrt2}(|0\rangle_2+|1\rangle_2)\).
- \(\CNOT_{2\to1}\) then produces the Bell state
\[
|\Phi^+\rangle_{2,1}
=\frac{1}{\sqrt2}\bigl(|0\rangle_2|0\rangle_1+|1\rangle_2|1\rangle_1\bigr)
\,,
\]
while qubits 0,3 remain \(|0\rangle_0|0\rangle_3\).
3. Next \(H_0\) on qubit 0 makes \(\tfrac1{\sqrt2}(|0\rangle_0+|1\rangle_0)\).
4. Finally \(\CNOT_{0\to3}\) on \((0,3)\) yields the Bell pair
\[
|\Phi^+\rangle_{0,3}
=\frac{1}{\sqrt2}\bigl(|0\rangle_0|0\rangle_3+|1\rangle_0|1\rangle_3\bigr)\,.
\]
Hence the ideal 4-qubit state is the product of two Bell states on disjoint pairs:
\[
|\Psi_{\rm ideal}\rangle
=|\Phi^+\rangle_{0,3}\;\otimes\;|\Phi^+\rangle_{2,1}\,.
\]
One checks by expanding that in the computational basis (ordering 0,1,2,3)
\[
|\Psi_{\rm ideal}\rangle
=\frac12\bigl|0000\bigr\rangle
+\frac12\bigl|0110\bigr\rangle
+\frac12\bigl|1001\bigr\rangle
+\frac12\bigl|1111\bigr\rangle
=\frac{|00_L\>+|11_L\>}{\sqrt2}\,,
\]
the logical two‐qubit GHZ in the [[4,2,2]] code.
Step 2: Noise model and its action on each Bell pair
Each \(\CNOT\) gate carries immediately afterwards a two‐qubit depolarizing channel on its two qubits:
\[
\Lambda(\rho)
=(1-p)\,\rho
\;+\;\frac{p}{15}\sum_{(i,j)\neq(I,I)}\bigl(\sigma_i\otimes\sigma_j\bigr)\,\rho\,
\bigl(\sigma_i\otimes\sigma_j\bigr)\,.
\]
– The first such channel \(\Lambda_{2,1}\) acts on the nascent Bell \(|\Phi^+\rangle_{2,1}\);
– the second \(\Lambda_{0,3}\) acts on \(|\Phi^+\rangle_{0,3}\).
Because these supports \(\{2,1\}\) and \(\{0,3\}\) are disjoint and subsequent gates on \(\{0,3\}\) do not spread errors onto \(\{1,2\}\), the two noisy Bell pairs remain uncorrelated. Thus the final state is
\[
\rho_{\rm final}
=\Bigl(\Lambda_{0,3}\bigl(|\Phi^+\>\!\<\Phi^+|\bigr)\Bigr)
\;\otimes\;
\Bigl(\Lambda_{2,1}\bigl(|\Phi^+\>\!\<\Phi^+|\bigr)\Bigr)\,.
\]
The fidelity of the 4-qubit state with its ideal \(|\Psi_{\rm ideal}\rangle
=|\Phi^+\>_{0,3}\otimes|\Phi^+\>_{2,1}\) then factorizes:
\[
F(p)
=\<\Phi^+|\,\Lambda_{0,3}(|\Phi^+\>\<\Phi^+|)\,|\Phi^+\>
\;\times\;
\<\Phi^+|\,\Lambda_{2,1}(|\Phi^+\>\<\Phi^+|)\,|\Phi^+\>
=\bigl[F_{\rm Bell}(p)\bigr]^2\,,
\]
where \(F_{\rm Bell}(p)\) is the fidelity of one Bell state under its two-qubit depolarizing channel.
Step 3: Fidelity of a Bell state under two-qubit depolarizing
Let \(\rho'=(1-p)\,|\Phi^+\>\<\Phi^+|+(p/15)\sum_{(i,j)\neq(I,I)}\sigma_i\otimes\sigma_j\,|\Phi^+\>\<\Phi^+|\,\sigma_i\otimes\sigma_j\). Then
\[
F_{\rm Bell}(p)
=\<\Phi^+|\rho'|\Phi^+\>
=(1-p)
\;+\;\frac{p}{15}
\sum_{(i,j)\neq(I,I)}
\bigl|\<\Phi^+|\sigma_i\otimes\sigma_j|\Phi^+\>\bigr|^2.
\]
One checks that the only nonzero overlaps occur for \(\sigma_i\otimes\sigma_i\) with \(i=X,Y,Z\), each giving \(\pm1\), so
\[
\sum_{(i,j)\neq(I,I)}
\bigl|\<\Phi^+|\sigma_i\otimes\sigma_j|\Phi^+\>\bigr|^2
=3.
\]
Hence
\[
F_{\rm Bell}(p)
=(1-p)+\frac{3p}{15}
=1-\frac{4}{5}\,p.
\]
Step 4: Global fidelity
Putting it all together,
\[
F(p)
=\Bigl(1-\tfrac{4}{5}p\Bigr)^2
=1-\tfrac{8}{5}p+\tfrac{16}{25}p^2.
\]
Final Answer:
\[
\boxed{F(p)\;=\;\biggl(1-\frac{4}{5}\,p\biggr)^2\,.}
\]