Suppose that $H$ and $P$ are $n \times n $ matrices such that $H$ is selfadjoint and $P$ is positive. Let $r \neq 0$ be a real number. Then consider the operator $A = H + irP$ - is it then invertible?
Notice that $A$ has real part $H$ and imaginary part $rP$ that means that if $A$ is normal which is equivalent to $\lbrack H, P \rbrack =0$ then the imaginary part of all of all eigenvalues are strictly positive or strictly negative which means that $\lambda = 0$ is not an eigenvalue and hence $A$ is invertible. But since $A$ is not nescessarily normal we can not conclude that. A nasty example is $A = \begin{pmatrix} 1 & -1 \newline 1 & -1 \end{pmatrix}$ which has $0$ as the only eigenvalue, but where $\frac{A+A^*}{2}$ has eigenvalues $1,-1$ and $\frac{A -A^*}{2}$ has eigenvalues $i,-i$. So they are both invertible, but the sum is not. I think one could find a similar example but I could not figure it out.
A matrix of the form $H + irP$, where $H$ and $P$ are Hermitian and $P \succ 0$, and where $0 \neq r \in \mathbb{R}$, has its numerical range $\{z \in \mathbb{C} \mid z = x^*(H + irP)x, \; x^*x = 1\}$ confined in the upper or lower half plane of $\mathbb{C}$, depending on the sign of $r$. Therefore, all its eigenvalues are also contained in the upper or lower half plane, and thus it is invertible.
Another way to see why it is invertible is to use simultaneous diagonalization of $H$ and $P$, which is possible since $P \succ 0$. For more on this, see, e.g., section 7.6. in the 2013 version of the book Matrix Analysis by Horn & Johnson.
Finally, you might be interested in the following paper by F. Zhang https://doi.org/10.1080/03081087.2014.933219 In particular, by multiplying your matrix with $\pm i$, depending on the sign of $r$, you obtain a matrix where the Hermitian part is positive definite.