Let $(x_{n}^{*})_{n}$ be a weak*-null sequence in $X^{*}$. The following are equivalent:
(1)$\sup\limits_{x\in K}|\langle x^{*}_{n},x\rangle|\rightarrow 0$ for each weakly compact subset $K$ in $X$;
(2)$|\langle x^{*}_{n},x_{n}\rangle|\rightarrow 0$ for each weakly null sequence $(x_{n})_{n}$ in $X$.
Indeed, if (1) is false, there exist a subsequence $(x^{*}_{k_{n}})_{n}$ of $(x^{*}_{n})_{n}$, a sequence $(x_{n})_{n}$ in $K$ and $\epsilon_{0}>0$ so that $|\langle x^{*}_{k_{n}},x_{n}\rangle|>\epsilon_{0}$ for all $n$. Since $K$ is weakly compact, there is a subsequence $(x_{n_{m}})_{m}$ of $(x_{n})_{n}$ that converges weakly to $x\in K$. Let us define a weakly null sequence $(z_{n})_{n}$ in $X$ by $z_{k_{n_{m}}}=x_{n_{m}}-x$ and $z_{n}=0$ otherwise. By (2), $\langle x^{*}_{n},z_{n}\rangle\rightarrow 0$. Note that $\langle x^{*}_{n},x\rangle\rightarrow 0$. This implies that $\langle x^{*}_{k_{n_{m}}},x_{n_{m}}\rangle\rightarrow 0$, a contradiction.
It follows from the above fact that my property (K) is equivalent to the Kalton-Pelczynski version of property (K).