Whether Fourier transform formula be considered as Convolution or Correlation? The expression for Fourier transform is given as
$$ F(\omega) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt  \tag{1}$$
Now, let one function be $f(x)$ and other be $e^{j\omega t}$ then their convolution at $p(0)$ is given as
$$ p(0) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt  \tag{2}$$ same as in $(1)$.
Also, if we split the exponential term into sinusoidal,the expression $(1)$ becomes
$$ F(\omega) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t} -j f(t)\sin{\omega t})\ dt  \tag{3}$$
Let $f(t)$ and $\cos{\omega t}$ be 2 functions. So their correlation  at value
$q(0)$  is given as
$$ q(0) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t}) \ dt  \tag{4}$$ same as in $(3)$. (we can do with f(t) and sine function also.)
So I want to know


Whether Fourier transform formula be considered as Convolution or Correlation?

Also, what is meaning of $p(0)$ and $q(0)$ terms ?
Correlation and convolution are basically the same operations. You can express the cross-correlation of two functions $f(t)$ and $g(t)$ by a convolution:

$$R_{fg}(\tau)=f(\tau)\star g^*(-\tau)$$

where $\star$ denotes convolution, and $*$ denotes complex conjugation.

If you evaluate the cross-correlation at $\tau=0$ you get the inner product of $f(t)$ and $g(t)$, and that's exactly what the Fourier transform is: it is the projection of $f(t)$ on the complex exponential $e^{j\omega t}$.

So, to answer your questions:


It is actually an inner product. Due to the equivalence of convolution and correlation, it can be seen as both, evaluated at $\tau=0$.
As already mentioned, the meaning of these terms is an inner product.