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 $\gamma$.