Let $K$ be a finitely generated extension of $\mathbb{Q}$ of transcendence degree at least 1.
Recall that a valuation ring of $K/\mathbb{Q}$ is a sub-$\mathbb{Q}$-algebra $V\subset K$ such that for every $x\in K$ we have $x\in V$ or $x^{-1}\in V$. A place of $K/\mathbb{Q}$ is a $\mathbb{Q}$-algebra homomorphism $p\colon V\to L$, where $V$ is a valuation ring of $K/\mathbb{Q}$, $L$ is a field and $p(x^{-1})=0$ for any $x\in K, x\not\in V$. The image of $p$ is an extension of $\mathbb{Q}$, which is called the residue field of $p$.
If $X/\mathbb{Q}$ is a projective variety with a $K$-rational point and $L$ is the residue field of a place of $K/\mathbb{Q}$, then it is easy to see that $X$ admits an $L$-rational point.
My question: Assume now that we are given a projective variety $Y/\mathbb{Q}$ that does not admit a $K$-rational point. Is there always a place $p$ of $K/\mathbb{Q}$ with residue field $L$, where $\operatorname{trdeg}_{\mathbb{Q}} L < \operatorname{trdeg}_{\mathbb{Q}} K$, such that $Y$ also does not admit an $L$-rational point?
