Satisfying the following determinant inequality I would like to find least restrictive conditions on $W = W^T \succ 0, \ V = V^T \succ 0$ (which are $\mathbb{R}^{n \times n}$ positive definite matrices) such that the following inequality is satisfied: 
$$ \text{det} \bigg( W^{-1} \Gamma W^{-1} + A^T V A \bigg) \geq 1  \tag{*}  $$
where $\Gamma = \Gamma^T \triangleq W - B P B^T \in \mathbb{R}^{n \times n} $, but is not necessarily a positive definite matrix, with $P = P^T \succ 0$ being a $\mathbb{R}^{n \times n}$ positive definite matrix. Further, the matrices $A, B \in \mathbb{R}^{n \times n}$ are arbitrary. 
I was hoping that (*) can be simplified by using appropriate determinant inequalities, for example using Minkowski's determinant inequality (here: https://mathoverflow.net/questions/251646/reverse-minkowski-and-related-determinant-inequalities), but this requires that $\Gamma$ be positive definite--though it is symmetric $\Gamma = \Gamma^T$--so not sure. Further, the second term would need to be positive definite as well. I would not want to impose conditions on $A,B$ though.
 A: I have listed some observations below.
What you want appears to be impossible without restricting $A$ and $B$ heavily.
First observation:
If $n$ is odd, there are no positive definite matrices $W,V,P$ so that (*) holds for all $A,B$.
Consider $W,V,P,A$ fixed and let $B=tI$ for $t\in\mathbb R$.
Then the matrix whose determinant you are interested in is
$$
X-t^2Y,
$$
where $Y=W^{-1}PW^{-1}$ and $X=W^{-1}+A^TVA$.
Notice that $X$ and $Y$ are independent of $t$.
When $t$ is large, the term $-t^2Y$ dominates, and you can essentially forget $X$.
More rigorously1,
$$
\det(X-t^2Y)
=
\det(-t^2Y)+O(t^{2n-2})
=
(-1)^nt^{2n}\det(Y)+O(t^{2n-2}).
$$
Notice that $\det(Y)=\det(P)\det(W)^{-2}>0$.
The exact form of the error term is irrelevant; it's enough that it's $O(t^{2n-2})$ so that it grows slower than the first term.
In fact, $t\mapsto\det(X-t^2Y)$ is a polynomial, and the sign of a polynomial for large argument only depends on the leading term.
What all this means is that if $n$ is odd, then the determinant will be negative for large enough $t$.
In particular, it cannot be at least 1 for all $t$.
It was not important that $B$ is a multiple of the identity.
The same argument goes through for $B=tC$ for any invertible $C$; you would just have $Y=W^{-1}BPB^TW^{-1}$ which has positive determinant.
The conclusion is: The desired identity cannot hold for all $B$ if $n$ is odd, no matter what you do.
(To prove that something doesn't hold for all $B$, it's enough to find one counterexample. I chose the simplest one, but there are many more.)
Second observation:
If $A=B=0$, then your (*) becomes $\det(W)\leq1$.
This is a necessary assumption to make on $W$.
I don't know if it's sufficient in even dimensions.
Third observation:
The determinant is independent of basis.
You can use this freedom to diagonalize one of your symmetric matrices.
Fourth observation:
Any positive definite matrix has a well defined positive definite square root.
(To find it: diagonalize and take the root of each diagonal element.)
There is a symmetric matrix $W^{-1/2}\in\mathbb R^{n\times n}$ which is positive definite and whose square is $W^{-1}$.
Similarly, there is a symmetric $P^{-1/2}$.
Now choose $B=W^{-1/2}P^{-1/2}$.
The transpose is $B^T=P^{-1/2}W^{-1/2}$.
Then it happens that
$$
\Gamma
=
W-W^{-1/2}P^{-1/2}PP^{-1/2}W^{-1/2}
=
W-W^{-1/2}IW^{-1/2}
=
0.
$$
Now if you also choose $A=0$, then $W^{-1} \Gamma W^{-1} + A^T V A=0$ and (*) fails.
This works for any positive definite $W,V,P$.
Fifth observation:
If you are willing to restrict $A$ and $B$, there is a way.
Starting with observation 2, assume $\det(W)<1$.
Let $Z=A^TVA-W^{-1}BPB^TW^{-1}$.
We have $\det(W^{-1}+Z)=\det(W)^{-1}\det(I+WZ)$.
Denote by $\|Q\|_\infty$ the largest absolute value of the elements of a matrix, and by $\|Q\|$ the operator norm.
Suppose $\|Q\|_\infty\leq1$.
Using the polynomial formula
$$
\det(Q)= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} Q_{\sigma(i),i}
$$
for a determinant, one can see that
$$
\begin{split}
&
|\det(I+Q)-1|
\\&=
\left|
\prod_{i=1}^{n} (1+Q_{i,i})-1
+
\sum_{\sigma\neq\text{id}} \text{sgn}(\sigma) \prod_{i=1}^{n} (\delta_{\sigma(i),i}+Q_{\sigma(i),i})
\right|
\\&\leq
\left|
\prod_{i=1}^{n} (1+Q_{i,i})-1
\right|
+
\sum_{\sigma\neq\text{id}}
\left|
\prod_{i=1}^{n} (\delta_{\sigma(i),i}+Q_{\sigma(i),i})
\right|
\\&\leq
(2^n-1)\times\|Q\|_\infty
+
(n!-1)\times 2^{n-1}\|Q\|_\infty
\\&\leq
2^nn!\|Q\|_\infty.
\end{split}
$$
Each term in the product $\left|
\prod_{i=1}^{n} (\delta_{\sigma(i),i}+Q_{\sigma(i),i})
\right|$ above is at most $2$ in absolute value, and there is at least one $i$ so that $\delta_{\sigma(i),i}=0$, so the product is at most $2^{n-1}\|Q\|_\infty$ in absolute value.
The rest of the estimates are similar.
If $A$ and $B$ are such that $\|Z\|<\epsilon$, then $\|WZ\|_\infty\leq\|WZ\|\leq\|W\|\|Z\|<\epsilon\|W\|$, and so
$$
\begin{split}
\det(W^{-1}+Z)
&=
\det(W)^{-1}\det(I+WZ)
\\&=
\det(W)^{-1}
-
\det(W)^{-1}(\det(I+WZ)-1)
\\&\geq
\det(W)^{-1}
-
\det(W)^{-1}2^nn!\|WZ\|_\infty
\\&\geq
\det(W)^{-1}
-
2^nn!\|W\|\epsilon\det(W)^{-1}.
\end{split}
$$
Now if $\epsilon\leq\frac{1-\det(W)}{2^nn!\|W\|}$, then $\det(W)^{-1}
-
2^nn!\|W\|\epsilon\det(W)^{-1}\geq1$
and so (*) holds.
To get $\|Z\|<\epsilon$, it is enough that $\|A\|<\sqrt{\epsilon/2\|V\|}$ and $\|B\|<\|W^{-1}\|^{-1}\sqrt{\epsilon/2\|P\|}$; the estimate leading to this is immediate from the formula for $Z$.
This leads to explicit restrictions on $A$ and $B$.
Using the $\epsilon$ from above, we get
$$
\|A\|<\min\left(\sqrt{\frac{1-\det(W)}{2^{n+1}n!\|W\|\|V\|}},\sqrt{1/2\|V\|}\right)
$$
and
$$
\|B\|<\min\left(\|W^{-1}\|^{-1}\sqrt{\frac{1-\det(W)}{2^{n+1}n!\|W\|\|P\|}},\|W^{-1}\|^{-1}\sqrt{1/2\|P\|}\right).
$$
(I also made the assumption $\epsilon\leq1$ to simplify some calculations.)
If you need more details on this observation, please ask a separate question; it would be sidetrack here to go into all details, since your original question wanted no restrictions on $A$ and $B$.

1
The second inequality is nothing more than $\det(\lambda A)=\lambda^n\det(A)$.
The first one is a little harder to see.
One option is to rewrite it as
$$
\det(X-t^2P)
=
\det((P^{-1}X-t^2I)P)
=
\det(P^{-1}X-t^2I)\det(P).
$$
Now it's clear that it's a characteristic polynomial (with $t^2$ instead of $\lambda$) times $\det(P)$.
The leading order term of $\det(Q-\lambda I)$ is $(-\lambda)^n$, and it is a polynomial of degree $n$ in $\lambda$.
Substitute $\lambda=t^2$ to get the result I used above.
