This limit doesn't exist; the proportion oscillates because the highest numbers dominate it. For instance, for $p_n\approx5\cdot10^k$, most numbers will be less than their reversed versions, simply because they start with a digit from $1$ to $4$, whereas for $p_n\approx10^k$ the proportion will be roughly $\frac12$.
Instead, you should consider the logarithmic density, which weights each range $\left[10^n,10^{n+1}\right]$ equally.
By the prime number theorem for arithmetic progressions, the primes are asymptotically equidistributed over the admissible residue classes mod $10^k$. Thus, $\frac14$ of them end in each of $1,3,7,9$, the ones ending in $1$ are equidistributed with respect to ending in $01,11,\ldots,91$, and so on. Thus, the reversed version of a prime is effectively a random number with the first digit uniformly randomly drawn from $1,3,7,9$ and the following digits uniformly randomly drawn from all digits. A number $x$ distributed in $\left[10^n,10^{n+1}\right]$ with density $\frac1{x\log10}$ thus has the following probability to be less than its reversed version:
\begin{eqnarray}
&&
\frac1{4\log10}\left(\int_1^2\frac{2-x}x\mathrm dx+\log3+\int_3^4\frac{4-x}x\mathrm dx\\+\log7+\int_7^8\frac{8-x}x\mathrm dx+\log9+\int_9^{10}\frac{10-x}x\mathrm dx\right)
\\
&=&
\frac{2\log2+\log7+8\log2-4\log3+24\log2-8\log7+\log9+9\log10-18\log3-4}{4\log10}
\\
&=&
\frac{43\log2-20\log3+9\log5-7\log7-4}{4\log10}
\\
&\approx&
0.5099\;.
\end{eqnarray}
This is the logarithmic density of the primes that are less than their reversed versions.
