Your both conjectural inequalities are equivalent to each other and false even for $n=p$, $A=B$, $X=I$ (and $a_i=\lambda^i$ for large $\lambda$, for example).

What is true that $$|X'AX+B|\geqslant c(X)\prod_{i=1}^p (a_{n-p+i}+b_i).$$

**Proof.** Let $\lambda_1\geqslant \lambda_{2}\geqslant \ldots \geqslant \lambda_p$ be eigenvalues of $X'AX+B$. It suffices to prove that $$\lambda_{i}\geqslant \rho(b_i+a_{n-p+i})\quad (*)$$ for all $i=1,\dots,p$, where $\rho=\rho(X)$ depends only on $X$.
Consider two cases.

1) $b_i\geqslant a_{n-p+i}$. Then $(*)$ holds with $\rho=1/2$, since $X'AX+B\geqslant B$, thus $\lambda_i\geqslant b_i$.

2) $a_{n-p+i}\geqslant b_i$. Let $L$ be the image of the operator (identified with the matrix) $X$, $\dim L=p$ and $X$ is a linear isomorphism onto $L$. Denote by $4\rho^2$, $\rho>0$, the norm of the inverse map $X^{-1}:L\mapsto X$. The intersection $L_i$ of $L$ and the space generated by $u_1,\dots,u_{n-p+i}$, where $Au_i=a_i u_i$ and $u_i$ are orthogonal eigenvectors of $A$, has dimension at least $i$. We have $$((X'AX+B) x,x)\geqslant (X'AX x,x)=(AXx,Xx)\geqslant a_{n-p+i}(Xx,Xx)\geqslant 2\rho a_{n-p+i}(x,x)$$ whenever $x\in X^{-1}L_i.$ Therefore by the variational (Courant) principle $\lambda_i\geqslant 2\rho(X)a_{n-p+i}\geqslant \rho(a_{n-p+i}+b_i)$.