Fix a commutative ring $R$. Lurie proved that smooth proper $R$-linear stable infinity categories are dualizable in $\text{Cat}^\text{perf}_{R,\infty}$.
For a smooth proper variety $X$ over $R$, what is the dual of the category of perfect complexes on $X$ in $\text{Cat}^\text{perf}_{R,\infty}$?
It is self-dual. In general the dual of a smooth proper $R$-linear $\infty$-category $C$ is always $C^{\operatorname{op}}$, but for a scheme $X$ we have $\operatorname{Perf}(X) = \operatorname{Perf}(X)^{\operatorname{op}}$ via the monoidal duality.
An alternative perspective is the following: For a general qcqs scheme $X$ over $R$, $\operatorname{QCoh}(X)$ is self-dual in $\operatorname{Pr}^{\operatorname{L}}_R$ via the pairing $(\mathcal F,\mathcal G) \mapsto \Gamma(\mathcal F\otimes \mathcal G)$ (in other words, $\operatorname{QCoh}(X)$ is a commutative Frobenius algebra in $\operatorname{Pr}^{\operatorname{L}}_R$, see SAG Corollary 9.4.3.4). The assumption that $X$ is smooth resp. proper over $R$ guarantees that the coevaluation resp. the evaluation preserves perfect complexes, so when it is smooth and proper the self-duality restricts to one in $\operatorname{Cat}^{\operatorname{perf}}_R$.
ETA: Let me be more precise about the last sentence, since the geometric properties "smooth" and "proper" by themselves do not exactly correspond to the categorical ones. For the global section functor $\Gamma\colon \operatorname{QCoh}(X)\to \operatorname{Mod}_R$ to preserve perfect complexes, one needs $X$ to be proper, of finite presentation and of finite Tor-amplitude. On the other hand, the coevaluation is given by pushing forward the structure sheaf along the diagonal $\delta\colon X\to X×_R X$. If $X$ is smooth and separated over $R$ then $\delta$ is proper, of finite presentation and of finite Tor-amplitude, so that $\delta_*$ preserves perfect complexes.
