This is not really an answer, but an observation too long for a comment. The OP may already know this ...

From the solution of Lyapunov equation $P = \int_{0}^{\infty} e^{At}Xe^{A^{\top}t} {\mathrm{d}}t$, we have ${\rm{tr}}(PX) = \int_{0}^{\infty} {\rm{tr}}(YX){\mathrm{d}}t$, where $Y := e^{At}Xe^{A^{\top}t} \succeq 0$.

One can then get a bound by using the fact that for positive semi-definite $Y,X$, we have $\lambda_{\min}(Y){\rm{tr}}(X) \leq {\rm{tr}}(YX) \leq \lambda_{\max}(Y){\rm{tr}}(X)$. Using this equality with ${\rm{tr}}(X)=1$ gives an upper bound:

$$ {\rm{tr}}(PX) \leq \int_{0}^{\infty} \parallel e^{At} X^{1/2}\parallel_{2}^{2} {\rm{d}}t \leq \lambda_{\max}(X)\int_{0}^{\infty} \parallel e^{At}\parallel_{2}^{2} {\rm{d}}t$$

where the last step used sub-multiplicative property of 2-norm. Not sure how tight is this.

This is not really an answer, but an observation too long for a comment. The OP may already know this ...

From the solution of Lyapunov equation $P = \int_{0}^{\infty} e^{At}Xe^{A^{\top}t} {\mathrm{d}}t$, we have ${\rm{tr}}(PX) = \int_{0}^{\infty} {\rm{tr}}(YX){\mathrm{d}}t$, where $Y := e^{At}Xe^{A^{\top}t} \succeq 0$.

One can then get a bound by using the fact that for positive semi-definite $Y,X$, we have $\lambda_{\min}(Y){\rm{tr}}(X) \leq {\rm{tr}}(YX) \leq \lambda_{\max}(Y){\rm{tr}}(X)$. Using this inequality with ${\rm{tr}}(X)=1$ gives an upper bound:

$$ {\rm{tr}}(PX) \leq \int_{0}^{\infty} \parallel e^{At} X^{1/2}\parallel_{2}^{2} {\rm{d}}t \leq \lambda_{\max}(X)\int_{0}^{\infty} \parallel e^{At}\parallel_{2}^{2} {\rm{d}}t$$

where the last step used sub-multiplicative property of 2-norm. Not sure how tight is this.