For single particles, S orbitals have angular momentum quantum number $\ell = 0$, and P orbitals have $\ell = 1$. The magnetic quantum number $m$ runs from $-\ell$ to $+\ell$ in integer steps, so S orbitals can only have one value of $m=0$ (hence a singlet state), and P orbitals can have $m=-1,0,1$ (hence a triplet of states). Is this what you are asking?
Edit
What I think your instructor meant is along the following lines. For a two-electron state, the combined wavefunction of the two electron system needs to be antisymmetric under exchange of electrons. The total wavefunction is the tensor product of the spin and spatial parts, so if one is symmetric, the other needs to be antisymmetric.
Note that the singlet $|\downarrow \uparrow\rangle - |\uparrow \downarrow\rangle$, with $s=0$, is antisymmetric under particle exchange, so the wavefunction of the electrons needs to be symmetric. So we can arrange $\Psi(x_1, x_2)$ for the spatial part of the wavefunction to be symmetric. The theory of addition of angular momentum tells us that states with even angular momentum are symmetric under exchange, so in this case $\ell = \ell_1 + \ell_2 = 0, 2, \dots$ is allowed. For the S state ($\ell = 0$), we can just write $\Psi(x_1, x_2) = \psi_s(x_1) \psi_s(x_2)$ where $\psi_s$ is a single-particle wave-function with $\ell_{1,2} = 0$.
For the triplet states, with $s=1$, each spin state is symmetric under particle exchange, so we need to arrange a spatial wavefunction that is antisymmetric. Allowable spatial wavefunctions have $\ell = 1, 3,\dots$. The simplest is $\ell=1$ which is a symmetrized linear combination of $\psi_s$ and $\psi_p$. The exact form of the wavefunction depends on $m_\ell$.
Note though that we are doing all of this from the point of view of the spins. So that while the $S$ state has both particles individually in an $S$ state, the two-particle $P$ state has individual particles in a superposition of $S$ and $P$ states.
