Here are the definitions for property $p$-$(V)$ and property $(V)$.
A Banach space $X$ has property $(V)$ if and only if every unconditionally converging operator $T$ from $X$ to any Banach space $Y$ is weakly compact.
A Banach space $X$ has property $(V)$ if every $V$-subset of $X^*$ is relatively weakly compact.
$C(K)$ spaces and reflexive spaces have property $(V)$.
A bounded subset $A$ of $X^*$ is called a $V$-subset of $X^*$ provided that $$ \sup_{x^* \in A }|x^* (x_n )| \to 0 $$ for each weakly unconditionally convergent series $\sum x_n$ in $X$.
A Banach space $X$ has property $p$-$(V)$ if for every Banach space $Y$, every $p$-converging operator $T:X\to Y$ is weakly compact.
A bounded subset $A$ of $X^*$ is called an $L$-subset of $X^*$ if for all weakly null sequences $(x_n)$ in $X$, $\sup_{x^*\in A} |x^*(x_n)|\to 0 $.
Let $1\le p\le \infty$. A bounded subset $A$ of $X^*$ is called a $p$-$(V)$ set if for all weakly $p$-summable (weakly null for $p=\infty$) sequences $(x_n)$ in $X$, $\sup_{x^*\in A} |x^*(x_n)|\to 0 $.
A Banach space $X$ has property $p$-$(V)$ if every $p$-$(V)$ susbset of $X^*$ is relatively weakly compact.
We note that the $1$-$(V)$ sets are the same as the $(V)$-sets and the $\infty$-$(V)$ sets are the same as the $L$-sets; property $1$-$(V)$ is precisely property $(V)$ and property $\infty$-$(V)$ is the $RDP$ property.
A Banach space $X$ has the reciprocal Dunford-Pettis ($RDP$) property if every $L$-subset of $X^*$ is relatively weakly compact. Equivalently, $X$ has the $RDP$ property if every completely continuous operator on $X$ is weakly compact.
I am wondering if there is an example of an $\mathcal{L}_\infty$ Banach space with property p-(V) and without property (V).
If X has property (V), then X has property p-(V) (this follows from definitions, since any p-convergent operator is unconditionally converging). Thus C(K) spaces are $\mathcal{L}_\infty$ spaces with properties $(V)$ and $p$-$(V)$.
Theorem 2.7 in [1] shows that the James $p$-space $J_p$ has property $p^*$-$(V)$ (here $p^*$ denotes the conjugate of $p$).
In the paper [1] it is shown that the James $p^*$-space $J_{p^*}$ has property $p$-$(V)$ and does not have property $(V)$ (p.421).
I don't think the James $p^*$-space $J_{p^*}$ is an $\mathcal{L}_\infty$ space.
[1] L. Li, D. Chen, and J. Alejandro Chavez-Dominguez, Pe{\l}czy'nski's Property $(V^*)$ of order $p$ and its quantification, Math. Nach. 291 (2018), 420---442.
