Assume that there exists some $\epsilon>0$ such that for every compact $C$ we have $$\limsup_{n} P^n(x,C) \le 1-\epsilon.$$ Then why does this imply that the sequence $\{ \bar{P_k}(x,\cdot):k\in \mathbb{N} \}$ is not tight, where we define $$\bar{P_k}(x,\cdot):= \frac{1}{k} \sum_{i=1}^k P^i(x,\cdot).$$
Tight here means that for each $\epsilon >0$ there exists a compact set $C$ such that $\liminf_{k\to \infty} \bar{P_k}(x,C) \ge 1-\epsilon$.
I think this uses the idea that we have for sufficiently large $n \ge N$, $P^n(x,C)\le 1-\epsilon /2$. And if the average of $P^i(x,\cdot)$ is tight, then we must have $$\liminf_k \frac{1}{k}\sum_{i=1}^k P^i(x,C) \ge 1-\epsilon/4$$ which implies that for some $i$, we must have $P^i(x,C) \ge 1-\epsilon/4$.
However, I cannot guarantee that this $i$ must be $\ge N$, to derive a contrdiction. I would greatly appreciate any help with this.
Well, $\limsup_{n} P^n(x,C) \le 1-\epsilon$ implies that for some $N$, one has $P^n(x,C) \le 1-\epsilon<1-\epsilon/2$ for all $n\ge N$, as you have already observed. Now for $k\geq N$, we have $$\frac{1}{k} \sum_{i=1}^k P^i(x,C)=\frac{1}{k}\sum_{i=1}^{N-1} P^i(x,C)+\frac{1}{k}\sum_{i=N}^{k} P^i(x,C)\le\frac{1}{k}\sum_{i=1}^{N-1} P^i(x,C)+\frac{k-N+1}{k}(1-\epsilon/2).$$ Since $$\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^{N-1} P^i(x,C)=0,$$ and $$\frac{k-N+1}{k}(1-\epsilon/2)\leq (1-\epsilon/2)$$ for all $k\geq N$ we get $$\limsup_k \frac{1}{k}\sum_{i=1}^k P^i(x,C) \leq 1-\epsilon/2,$$ and a fortiori also $$\liminf_k \frac{1}{k}\sum_{i=1}^k P^i(x,C) \leq 1-\epsilon/2.$$