Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer?

Because $$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} = a^2 + a b + a c + b^2 + b c + c^2 \text{,} $$ the result is certainly an integer for distinct integers $a$,$b$,$c$. (It's undefined if any pair of them are equal.)

Your guess about $P$ being the product of differences of alphabetical pairs is correct.

Up to a minus sign, $\overset{2}{C}(2,3,5) = -69 $, is the negative of the first example you gave.

The statement is true even when you replace the numerators by any polynomial with integer coefficients or increase the number of variables. More precisely,

For any $f(x) \in \mathbb{Z}[x]$ and distinct $a_1, \ldots, a_n \in \mathbb{Z}$, $\displaystyle\;\sum\limits_{k=1}^n \frac{f(a_k)}{\prod\limits_{j=1,\ne k}^n (a_k - a_j)}\;$ is an integer.

I will only prove the statement for $3$ distinct integers $a,b,c$. The proof for other $n$ is similar.

For any $f(x) \in \mathbb{Z}[x]$, long divide it by $(x-a)(x-b)(x-c)$.
This gives us two polynomials $g(x), h(x) \in \mathbb{Z}[x]$ such that

$$f(x) = (x-a)(x-b)(x-c)g(x) + h(x)$$

and $h(x) = Ax^2 + Bx + C$ is at most quadratic.

Apply partial fraction decomposition to $\displaystyle\;\frac{h(x)}{(x-a)(x-b)(x-c)}\;$ and notice $$f(a) = h(a),\quad f(b) = h(b)\quad\text{ and }\quad f(c) = h(c)$$

We obtain

$$\begin{align} \frac{h(x)}{(x-a)(x-b)(x-c)} &= \frac{1}{x-a}\frac{h(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{h(b)}{(b-a)(b-c)} + \frac{1}{x-c}\frac{h(c)}{(c-a)(c-b)}\\ &= \frac{1}{x-a}\frac{f(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{f(b)}{(b-a)(b-c)} + \frac{1}{x-c}\frac{f(c)}{(c-a)(c-b)} \end{align} $$ This implies $$\frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)} = \lim_{x\to\infty} \frac{xh(x)}{(x-a)(x-b)(x-c)} = A$$ The coefficients of $x^2$ in $h(x)$ which is an integer.

For the original problem, we can take $f(x) = x^4$ and conclude

$$\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \in \mathbb{Z} \quad\text{ for distinct }\; a,b,c \in \mathbb{Z} $$

According to Wolfram Alpha the expression

$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(b-a)}+\frac{c^4}{(c-a)(c-b)} \tag{$\ast$}$$ is not always an integer (see https://www.wolframalpha.com/input/?i=%5Cfrac%7Ba%5E4%7D%7B(b-a)(c-a)%7D%2B%5Cfrac%7Bb%5E4%7D%7B(c-b)(b-a)%7D%2B%5Cfrac%7Bc%5E4%7D%7B(c-a)(c-b)%7D,+a%3D3,b%3D5,c%3D7). But the expression you wrote

$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} \tag{$\ast$}$$ is always an integer. (Note the difference with your example. The denominator of the second fraction changes sign).

With respect to the determinant note that:

\begin{align}\left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\ 1 & d & d^2 & d^4 \\ \end{array} \right| & \\ &\underbrace{=}_{(1)}\left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\ 0 & b-a & b^2-a^2 & b^4-a^4 \\ 0 & c-b & c^2-b^2 & c^4-b^4 \\ 0 & d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right| \\ &\underbrace{=}_{(2)}\left|\begin{array}{ccc} b-a & b^2-a^2 & b^4-a^4 \\ c-b & c^2-b^2 & c^4-b^4 \\ d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right|\\ &\underbrace{=}_{(3)}(b-a)(c-b)(d-c)\left|\begin{array}{ccc} 1 & b+a & b^3+b^2a+ba^2+a^3 \\ 1 & c+b & c^3+c^2b+cb^2+b^3 \\ 1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| \end{align}

where in $(1)$ we have made $F_i-F_1\to F_i, i=2,3,4,$ in $(2)$ we have used the Laplacian's expansion using the first column and in $(3)$ we get out of the determinand $b-a$ from the first row, $c-b$ from the second one and $d-c$ from the third one.

Now, proceeding in a similar way we have \begin{align}\left|\begin{array}{ccc} 1 & b+a & b^3+b^2a+ba^2+a^3 \\ 1 & c+b & c^3+c^2b+cb^2+b^3 \\ 1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| &\\&=\left|\begin{array}{ccc} 1 & b+a & b^3+b^2a+ba^2+a^3 \\ 0 & c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\ 0 & d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=\left|\begin{array}{ccc} c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\ d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=(d-b)(d-a)(c-a)(a+b+c+d).\end{align}

So, your guess about $P$ is correct.