 Solution

Let $E_{\ell }$ be the event that the block beginning at position $\ell$, $1\le \ell \le k+1$, is a spalindrome. Each has probability $(10{)}_k/10^{2k}$. If two starting positions differ by $d<k$, the first half of the later block contains the middle digit of the earlier palindrome twice, contradicting distinctness. Thus the only possible overlap is $E_1\cap E_{k+1}$.

That intersection consists of words $A{A}^{\mathrm{rev}}A$, where $A$ has $k$ distinct digits, and hence has probability $(10{)}_k/10^{3k}$. This overlap structure of digit spalindromes makes inclusion--exclusion give