Step-by-Step Derivation
1. Ideal circuit and target state
We begin in |0000⟩ on qubits (0,1,2,3). The perfect (error‐free) gates
\[
H_2,\;\;CNOT_{2\to1},\;\;H_0,\;\;CNOT_{0\to3}
\]
produce the 4-qubit pure state
\[
|\psi\>\;=\;\Bigl(CNOT_{0\to3}\,H_0\,CNOT_{2\to1}\,H_2\Bigr)\,\ket{0000}
\;=\;|\Phi^+\>_{0,3}\;\otimes\;|\Phi^+\>_{2,1}\,,
\]
where
\[
|\Phi^+\> \;=\;\frac{\ket{00}+\ket{11}}{\sqrt2}\,.
\]
2. Error model
Each CNOT gate is followed immediately by a two-qubit depolarizing channel on its two qubits. Denote by \(\Lambda_{ij}\) the channel acting on qubits \(i,j\):
\[
\Lambda_{ij}(\rho)
=(1-p)\,\rho
\;+\;\frac{p}{15}\sum_{\substack{P\in\{I,X,Y,Z\}^{\!\otimes2}\\P\neq I\otimes I}}
P\,\rho\,P\,.
\]
No other errors are assumed.
Thus the actual final state is
\[
\rho
\;=\;
\Lambda_{0,3}\!\Bigl(\Lambda_{2,1}\bigl(|\psi\>\<\psi|\bigr)\Bigr)\,.
\]
3. Fidelity definition
The physical state fidelity is
\[
F(p)\;=\;\<\psi\,|\,\rho\,|\,\psi\>\;.
\]
4. Factorization over disjoint error channels
Since
\[
|\psi\>=|\Phi^+\>_{0,3}\otimes|\Phi^+\>_{2,1},
\]
and the channels \(\Lambda_{2,1}\) and \(\Lambda_{0,3}\) act on disjoint qubit pairs, the fidelity factorizes:
\[
F(p)
\;=\;
F_{2,1}(p)\;\times\;F_{0,3}(p),
\]
where
\[
F_{i,j}(p)
=\;\<\Phi^+|\;\Lambda_{i,j}\bigl(|\Phi^+\>\<\Phi^+|\bigr)\;|\Phi^+\>.
\]
5. Fidelity for one Bell pair under two-qubit depolarizing noise
Fix one pair, say qubits \((i,j)\), prepared in \(|\Phi^+\>\). Then
\[
\Lambda_{ij}(|\Phi^+\>\<\Phi^+|)
=(1-p)\,|\Phi^+\>\<\Phi^+|
\;+\;\frac{p}{15}\sum_{P\neq I\otimes I}P\,|\Phi^+\>\<\Phi^+|\,P.
\]
Hence
\[
F_{ij}(p)
=\<(1-p)\,|\Phi^+\>\<\Phi^+|
\;+\;\frac{p}{15}\sum_{P\neq I\otimes I}P\,|\Phi^+\>\<\Phi^+|\,P\>.
\]
Only those Pauli errors \(P\) in the set \(\{X\otimes X,\;Y\otimes Y,\;Z\otimes Z\}\) leave \(|\Phi^+\>\) invariant (up to phase), so
\[
\<\Phi^+|P|\Phi^+\> = 1
\quad\text{for }P\in\{XX,\,YY,\,ZZ\},
\]
and vanishes for the other 12 non-identity Paulis. Therefore
\[
F_{ij}(p)
= (1-p)\;+\;\frac{p}{15}\times 3
=1-p+\frac{3p}{15}
=1-\frac{4}{5}\,p.
\]
6. Total 4-qubit fidelity
Each of the two CNOTs contributes the same factor, so
\[
F(p)
= \Bigl(1-\tfrac{4}{5}\,p\Bigr)\times\Bigl(1-\tfrac{4}{5}\,p\Bigr)
=\bigl(1-\tfrac{4}{5}p\bigr)^{2}.
\]
Final Answer:
\[
F_{\rm phys}(p)\;=\;\bigl(1-\tfrac45\,p\bigr)^{2}\,.
\]