State whether the language is in $R$, $RE$, etc. The intuition for the solution
The intuition is that finite languages are very simple in the sense that throwing a finite number of words from a language does not affect membership in $\text{R}$ or in $\overline{\text{RE}}$, etc. Therefore, since $L^{\geq k}$ is the result of throwing a finite number of words from $L$, we get that:

$L(M)^{\geq k} \in \text{RE}$, and
$\overline{HP}^{\geq k} \notin \text{RE}$.

So the language $L = \{ \langle M \rangle: \exists k: \ L(M)^{\geq k} = \overline{HP}^{\geq k}\}$ must be empty.

Formally speaking, we have the following claims:
Claim 1: For every non-trivial language $A$, and every finite strict subset $B\subsetneq A$, it holds that $A \leq_m A \setminus B$.
Hint for the proof of claim 1: $B$ is finite and thus decidable. Therefore, given input $x$ for the reduction, we can check whether $x\in B$. Then you can proceed easily. Think where to map inputs $x$ from $B$, and where to map inputs $x$ from $\overline{B}$.
We also have the following similar claim which could be useful.
Claim 2: For every non-trivial language $A$, and every finite strict subset $B\subsetneq A$, it holds that $A\setminus B \leq_m A $.
Given the above claims, we're done. Indeed, $\overline{HP} \notin \text{RE}$, and for every $k$, $\overline{HP}^{\geq k} = \overline{HP}\setminus \{w\in \overline{HP} : |w| &lt; k\}$. That is, $\overline{HP}^{\geq k}$ equals a non-trivial infinite language minus a  finite subset, and so by claim 1, $\overline{HP}^{\geq k} \notin \text{RE}$. Also, it holds that $L(M)^{\geq k} \in RE$ for every machine $M$ (this is easy, you can prove that directly using standard closure properties and the fact that  $L(M)^{\geq k} = L(M)\setminus \{ w\in L(M): |w| &lt; k\}$. Alternatively, you can use claim 2 but you have to be careful regarding the edge cases where $L(M)$ is trivial, etc.). Therefore, it cannot be the case that there is a machine $M$ with $L(M)^{\geq k} = \overline{HP}^{\geq k}$.