Let us find a point $(x_0, y_0)$ that lies on all parabolae $P_a$ for any real $a$.

The equation of $P_a$ is $y = x^2 + a x + (2014 - a)$.

Let us try to find $x_0$ and $y_0$ such that for all $a$, $y_0 = x_0^2 + a x_0 + (2014 - a)$.

Rewriting:
$$y_0 = x_0^2 + a x_0 + 2014 - a$$
$$y_0 = x_0^2 + 2014 + a(x_0 - 1)$$
For this to be independent of $a$, the coefficient of $a$ must be zero:
$$x_0 - 1 = 0 \implies x_0 = 1$$
So, for $x_0 = 1$:
$$y_0 = (1)^2 + a(1) + 2014 - a = 1 + a + 2014 - a = 1 + 2014 = 2015$$
Therefore, the point $(1, 2015)$ lies on all parabolae $P_a$.

Thus, all parabolae $P_a$ pass through the same point $(1, 2015)$.