You have an irreducible, aperiodic Markov chain on the
set of all possible orders of books. Markov chain theory guarantees that
your limit exists and equals $\pi(123)$, where $\pi$ is the unique invariant
probability vector for the chain. In your notation, this turns out to be
$$\pi(123)={a_1\over a_1+a_2+a_3}\,{a_2\over a_2+a_3}\,{a_3\over a_3}.$$
For an arbitrary ordering of books, we get the corresponding result
$$\pi(ijk)={a_i\over a_i+a_j+a_k}\,{a_j\over a_j+a_k}\,{a_k\over a_k}.$$
A proof of this formula follows below.
More generally, let's denote the books as $A,B,C,\dots,Y,Z$ with
corresponding probability of being chosen as $a,b,c,\dots,y,z$.
Of course, $a+b+c+\cdots+y+z=1$.
Let's take a particular ordering, that I will label as $\beta$:
$$\beta:=DJG\cdots CW.$$ The formula for the invariant probability of
$\beta$ is the product
$$\pi(\beta):={d\over d+j+g+\cdots+c+w}\,{j\over j+g+\cdots+c+w}\,
{g\over g+\cdots+c+w} \cdots {c\over c+w}\, {w\over w}.$$
To check this, let's calculate the $\beta$th entry in the
vector $\pi P$, that is, $\sum_\alpha \pi(\alpha) P(\alpha,\beta)$.
The only states $\alpha$ from which it is possible to jump to state
$\beta$ are those that look just like $\beta$ but with book $D$ moved.
That is, $$\alpha\in\{ DJG\cdots CW, JDG\cdots CW, JGD\cdots CW, \dots,
JG\cdots CWD\}.$$ For all such $\alpha$, the transition probability
is simply $P(\alpha,\beta)=d$.
Therefore
$$
\begin{eqnarray*}(\pi P)(\beta)&=&d\sum_\alpha \pi(\alpha)\\
&=&{d\over d+j+g+\cdots+c+w} \sum_\alpha \pi(\alpha)\\
&=&{d\over d+j+g+\cdots+c+w}\, {j\over j+g+\cdots+c+w}\,
{g\over g+\cdots+c+w} \cdots {c\over c+w}\, {w\over w},
\end{eqnarray*}
$$
where, in the final step, we use the algebraic identity proved here.
The fact that $\sum_\gamma \pi(\gamma)=1$ can be proved by induction
or using the following interpretation: $\pi(\gamma)$ is the probability
that if I sample the books one at a time without replacement, they'd
occur in the order $\gamma$.