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Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.

Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset" (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.

EDIT : I should have said that $V$ is of dimension 2 here.

1

Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.

Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset" (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.