We know that electron posses dual behaviour and we consider them as stationary waves around a nucleus but how these stationary waves can exist in different orbitals, we can imagine circular stationary waves in the s orbital but when it comes to p orbital how do we even fit a stationary wave in the two lobes?
And is the electron actually orbiting around a nucleus in the non-spherical lobes, but how can it do so?
It's best to avoid the word "actually" when thinking about quantum mechnics. It implies some sort of underlying "reality" that looks like our familar world. We now know that there is no notion of "local reality" that is consistent with quantum mechanics.
And is the electron actually orbiting around a nucleus in the non-spherical lobes, but how can it do so?
To provide a meaningful answer one has to define more clearly what one means by "actually orbiting."
If we take the term "orbiting" to mean that the electron has a well defined trajectory (position as a function of time) then the answer is "no." This is basic quantum mechanics. The unfortunately-named "orbital" does not tell you the trajectory of an electron. It tells you the (square root of the) probability density to find the electron at a given position were you to measure its position: $$ dP(\vec r) = |\psi_{n\ell m}(\vec r)|^2d^3r\;, $$ where $$ \int dP = 1\;. $$
It we take the term "orbiting" to simply mean that the electron is described via an "orbital" like a p-orbital, then yes it is "orbiting" the nucleus (when the nucleus is assumed fixed), by definition. But this "orbiting" is not a classical trajectory.
how do we even fit a stationary wave in the two lobes?
The wave function is called "stationary" because its time dependence is relatively trivial: $$ \Psi_{n\ell m}(\vec r, t) = e^{-i E_{n}t}\psi_{n\ell m}(\vec r)\;,\tag{1} $$ where $\Psi_{n\ell m}(\vec r, t)$ is the solution to the full time-dependent Schrodinger equation, and $\psi_{n\ell m}(\vec r)$ (the "orbital") is the solution to the time-independent Schrodinger equation.
Because of the specific form of time dependence in Eq. (1), the (generally time-dependent) probability density turns out to be time independent, and is equal at all times to the probability density given by the orbital squared: $$ dP_{tdse}(\vec r, t)=|\Psi_{n\ell m}(\vec r, t)|^2 = |\psi_{n\ell m}(\vec r)|^2 = dP(\vec r)\;. $$
"Does an electron actually orbit around a nucleus?"
No, the electron and nucleus orbit around the barycenter. Remember, we solve the Coulomb atom in reduced radial coordinates:
$$ \vec r = \vec r_e - \vec r_N $$
with reduced mass:
$$ \mu = \frac 1 {\frac 1 {m_e} + \frac 1 {m_N}} $$
" how do we even fit a stationary wave in the two lobes?"
The angular part of the P orbit has the maximal $|m|$ part going as:
$$ Y_1^{\pm 1}(\theta, \phi) \propto \sin\theta e^{\pm\phi} $$
which has two lobes in the azimuth coordinate.
With the time evolution, the phase part looks like:
$$ e^{\pm i(\phi - E_n t/\hbar)} $$
which is a wave going around (counter)clockwise.
Of course, the probably density, $\psi^*\psi$, is constant--these are stationary states, the probability current:
$$ \vec j = \frac{\hbar}{2mi} \big( \psi^*\nabla\psi - \psi\nabla\psi^* \big) $$
orbits the nucleus as expected. (See the video linked in the comments)
"And is the electron actually orbiting around a nucleus in the non-spherical lobes, but how can it do so?"
The probably current certainly orbits. The electron and nucleus are just in an eigenstate.
Whether an atom really is in that state is dubious, since we made up the z-axis and didn't tell the atom. The thing about spherical harmonics at fixed $n$: the Coulomb solutions are all degenerate, so you can mix and match different $m$ states at fixed $l$ and get any axis you want. (You can also mix $l$ states, but there's no reason to do that--unless you're using parabolic coordinates, but I digress).
