= Solution
A \b[<state on a C-star algebra>] is a positive <linear functional> of <norm> one. In the unital case this is equivalent to positivity and $x(1)=1$. A unital star-homomorphism $\pi:A\to\mathcal B(K)$ is a \b[<cyclic representation of a C-star algebra>] if it has a <vector> $\xi$ with $\overline{\pi(A)\xi}=K$. For the representation associated with a state, the cyclic <vector> is taken to have <norm> one.
Use <inner products> linear in the first argument, as in the displayed formula for the projection in the source. In the \b[<Gelfand–Naimark–Segal construction>], set
$$
N_x=\{a\in A:x(a^*a)=0\},\qquad
\langle[a],[b]\rangle=x(b^*a).
$$
The <Cauchy–Schwarz inequality for positive C-star functionals> makes this <inner product> well-defined on $A/N_x$. Indeed a null <vector> pairs to zero with every <vector>. The set $N_x$ is a <left ideal>: $c^*c\leq\|c\|^21$ implies
$$
x((ca)^*(ca))\leq\|c\|^2x(a^*a),
$$
so $a\in N_x$ implies $ca\in N_x$. The same estimate shows that left multiplication induces a <bounded operator>
$$
\pi_x(c)[a]=[ca],\qquad\|\pi_x(c)\|\leq\|c\|.
$$
Complete the quotient pre-Hilbert space to $H_x$, and extend these operators by <continuity>. The formula for the <inner product> gives $\pi_x(c)^*=\pi_x(c^*)$, while multiplication and the unit are preserved. Put $\xi_x=[1]$. Then $\|\xi_x\|=1$, the <vectors> $\pi_x(a)\xi_x=[a]$ are dense in $H_x$, and
$$
\boxed{x(a)=\langle\pi_x(a)\xi_x,\xi_x\rangle.}
$$
This gives the requested cyclic representation for an arbitrary state.
For the specified unit <vector> $\eta$, the functional $x_\eta$ is complex-linear and obeys $|x_\eta(a)|\leq\|a\eta\|\|\eta\|\leq\|a\|$. Moreover
$$
x_\eta(1)=1,\qquad x_\eta(a^*a)=\|a\eta\|^2\geq0.
$$
Thus it is positive and has <norm> exactly one, proving that \b[$x_\eta$ is a <vector state on a C-star algebra>].
The rank-one projection satisfies
$$
(ap_\eta)(\zeta)=\langle\zeta,\eta\rangle a\eta.
$$
If $a\eta=0$, this is zero for every $\zeta$, so $ap_\eta=0$. Conversely, evaluation at $\zeta=\eta$ gives $(ap_\eta)\eta=a\eta$. Hence
$$
\boxed{a\eta=0\quad\Longleftrightarrow\quad ap_\eta=0.}
$$
In particular $N_{x_\eta}=\{a:a\eta=0\}=\{a:ap_\eta=0\}$.
The concrete GNS space is the <Hilbert space completion>[completion] of $A/N_{x_\eta}$ with
$$
\langle[a],[b]\rangle=x_\eta(b^*a)=\langle a\eta,b\eta\rangle.
$$
Define $U_0[a]=a\eta$. The description of the null ideal proves it is well-defined, and the displayed equality proves it is a <linear isometry> with range $A\eta$. It therefore extends to an <isometry> $U:H_{x_\eta}\to\overline{A\eta}$. The extended range is closed, since its domain is complete, and contains the dense set $A\eta$. Thus \b[$U$ is onto the required <closure>].
In fact here $A=\mathcal B(H)$, so $A\eta=H$: given any $\zeta\in H$, the bounded rank-one operator $b(v)=\langle v,\eta\rangle\zeta$ sends $\eta$ to $\zeta$. Thus $U$ is a unitary map onto all of $H$, and $U\xi_{x_\eta}=\eta$.
Finally, on the dense quotient <vectors>,
$$
U\pi_{x_\eta}(a)[b]=U[ab]=ab\eta=aU[b].
$$
Both sides are <bounded operators>, so this identity extends to all of $H_{x_\eta}$. Multiplication by the inverse unitary map gives
$$
\boxed{\pi_{x_\eta}(a)=U^{-1}aU\qquad(a\in A).}
$$
This is the <GNS representation of a vector state>, with the <vector> state, null ideal, cyclic <vector> and intertwining <isometry> all made explicit.
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