= Solution
First suppose $x$ is a <positive linear functional>, without assuming boundedness. Every <Hermitian C-star element> $h$ is a difference of positive elements, since $\|h\|1\pm h\geq0$. Thus $x(h)$ is real. For a <Hermitian C-star element> $h$ of <norm> at most one, $-1\leq h\leq1$ gives $|x(h)|\leq x(1)$.
For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The <real part> of $\zeta a$ is a <Hermitian C-star element> and has <norm> at most $\|a\|$. Since $x$ is real on <Hermitian C-star elements>, it respects real parts, giving
$$
|x(a)|=x(\operatorname{Re}(\zeta a))\leq x(1)\|a\|.
$$
This proves boundedness and $\|x\|\leq x(1)$. Evaluation at the unit gives the reverse inequality, so \b[positivity implies $\|x\|=x(1)$].
Conversely let $x$ be bounded with $x(1)=\|x\|=c\geq0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a <Hermitian C-star element> $h$ and real $t$, the <C-star identity> gives
$$
|1+itx(h)|\leq\|1+ith\|\leq\sqrt{1+t^2\|h\|^2}.
$$
Writing $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\leq t^2\|h\|^2$. Taking $t\to0$ through positive and negative values forces $b=0$. Hence $x$ is real on <Hermitian C-star elements>. If $0\leq a\leq1$, then $1-a$ has <norm> at most one, so $x(1-a)\leq1$ and $x(a)\geq0$. Scaling proves nonnegativity on every positive element, in particular every $b^*b$. Therefore
$$
\boxed{x\text{ positive}\iff\|x\|=x(1_A).}
$$
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