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ewcommand\Si{\Sigma}$ $

ewcommand\X{\mathbf X}$ The answer is no. Indeed, let $p_i:=P_i$, $p:=P$, $\X:=(X_1,\dots,X_n)$, and $\Si_\X:=\Si(p)$. At least one of the vectors $\Si^{-1}X_j$ is nonzero, for some $j$, because otherwise the matrix $$I=\Si_\X^{-1}\Si_\X=\sum_1^n p_i\Si_\X^{-1}X_i X_i^T$$ would be zero. So, there is some $j$ such that $c:=\|\Si_\X^{-1}X_j\|>0$. Replacing now $\X$ by $a\X$ for real $a>0$ and letting $a\to0$, we will have $$\|\Si_{a\X}^{-1}(aX_j)\|=\frac1a\,\|\Si_\X^{-1} X_j\|=\frac ca\to\infty,$$ so that the inequality $$\|\Si_{a\X}^{-1}(aX_j)\|\le\frac1{p_j}$$ will fail to hold for small enough $a$.

The OP has edited the question, thus invalidating this answer. However, even after the edit, the answer remains no. E.g., let $n=2$, $p_1=p_2=1/2$, $X_1:=[1,1]^T/\sqrt2$, and $X_2:=[1,1-h]^T/\sqrt2$ with $h\downarrow0$. Then $\|X_1\|=1$, $1\ge\|X_2\|\sim1$, but $$\|\Si_{\X}^{-1}(X_1)\|=\frac{2 \sqrt{2} \sqrt{h^2-2 h+2}}{h} \sim\frac4h

ot\lesssim2=\frac1{p_1}.$$