[guided]Fix $n\in\mathbb{N}$. From the preceding construction, define $p_n:=\varphi_a(n)$, define the extracted witness $m_n:=\pi_0(p_n)$, and define the accompanying realizer $r_n:=\pi_1(p_n)$. The existential realizability clause applied to $p_n\Vdash \exists m\,A(n,m)$ gives
\begin{align*}
r_n\Vdash A(n,m_n).
\end{align*}
Also define $c_n:=K(e,n)$, where $K:\mathbb{N}\times\mathbb{N}\rightharpoonup\mathbb{N}$ is the partial computable computation-code extractor from the theorem statement. Since $e=S(a)$ and $\varphi_e(n)=m_n$ halts, the defining property of $K$ gives that $T(e,n,c_n)$ is true and
\begin{align*}
U(c_n)=m_n.
\end{align*}
The target formula for this fixed $n$ is
\begin{align*}
\exists c\,\bigl(T(e,n,c)\land A(n,U(c))\bigr).
\end{align*}
To realize an existential statement, we must provide a coded pair whose first component is the witness $c$ and whose second component realizes the body at that witness. We choose the witness to be $c_n$.
It remains to realize the conjunction
\begin{align*}
T(e,n,c_n)\land A(n,U(c_n)).
\end{align*}
For the first conjunct, $T(e,n,c_n)$ is a true atomic arithmetic formula by the definition of $c_n$ as a halting computation code. Under the standard atomic clause for Kleene realizability, the fixed number $0$ realizes every true atomic formula. Hence
\begin{align*}
0\Vdash T(e,n,c_n).
\end{align*}
For the second conjunct, we have not directly extracted a realizer of $A(n,U(c_n))$; we extracted $r_n\Vdash A(n,m_n)$. The bridge is the equality $U(c_n)=m_n$, which is part of the defining property of the computation code. By substitution invariance of realizability under equal numerical values, the same realizer $r_n$ realizes the formula after replacing $m_n$ by the equal number $U(c_n)$:
\begin{align*}
r_n\Vdash A(n,U(c_n)).
\end{align*}
Now the conjunction clause says that the pair of these two realizers realizes the conjunction. Thus, with
\begin{align*}
d_n := \langle 0,r_n\rangle,
\end{align*}
we have
\begin{align*}
d_n\Vdash T(e,n,c_n)\land A(n,U(c_n)).
\end{align*}
Finally, pairing the witness $c_n$ with the conjunction realizer $d_n$ gives
\begin{align*}
q_n := \langle c_n,d_n\rangle,
\end{align*}
and the existential clause yields
\begin{align*}
q_n\Vdash \exists c\,\bigl(T(e,n,c)\land A(n,U(c))\bigr).
\end{align*}[/guided]