You can write the states $\Psi_{nljm_j}$ (which are eigenvectors of $J^2$ and $J_z$) as certain superpositions of the states $\Phi_{nlm_lm_s}$ (which are eigenvectors of $L_z$ and $S_z$). But unfortunately this angular momentum coupling is quite an elaborate task, and usually text books on quantum mechanics take at least one whole chapter to explain it. You have already figured out the number of configurations for the terms. The final result (with the coefficients taken from the table of Clebsch-Gordan coefficients, section 2.3) is this.

The term $^2P_{3/2}$ (i.e. with $l=1,j=\frac{3}{2}$) has $4$ configurations ($m_j=+\frac{3}{2},+\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$): $$\begin{align} \Psi_{21\frac{3}{2}+\frac{3}{2}} &= \Phi_{21+1\uparrow} \\ \Psi_{21\frac{3}{2}+\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21+1\downarrow} + \sqrt{\frac{2}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{3}{2}-\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{2}{3}}\Phi_{210\downarrow} \\ \Psi_{21\frac{3}{2}-\frac{3}{2}} &= \Phi_{21-1\downarrow} \end{align}$$

And the term $^2P_{1/2}$ (i.e. with $l=1,j=\frac{1}{2}$) has $2$ configurations ($m_j=+\frac{1}{2},-\frac{1}{2}$): $$\begin{align} \Psi_{21\frac{1}{2}+\frac{1}{2}} &= \sqrt{\frac{2}{3}}\Phi_{21+1\downarrow} - \sqrt{\frac{1}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{1}{2}-\frac{1}{2}} &= - \sqrt{\frac{2}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{1}{3}}\Phi_{210\downarrow} \end{align}$$

You can write the states $\Psi_{nljm_j}$ (which are eigenvectors of $L^2$, $S^2$, $J^2$ and $J_z$) as certain superpositions of the states $\Phi_{nlm_lm_s}$ (which are eigenvectors of $L^2$, $S^2$, $L_z$ and $S_z$). But unfortunately this angular momentum coupling is quite an elaborate task, and usually text books on quantum mechanics take at least one whole chapter to explain it. You have already figured out the number of configurations for the terms. The final result (with the coefficients taken from the table of Clebsch-Gordan coefficients, section 2.3) is this.

The term $^2P_{3/2}$ (i.e. with $s=\frac{1}{2},l=1,j=\frac{3}{2}$) has $4$ configurations ($m_j=+\frac{3}{2},+\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$): $$\begin{align} \Psi_{21\frac{3}{2}+\frac{3}{2}} &= \Phi_{21+1\uparrow} \\ \Psi_{21\frac{3}{2}+\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21+1\downarrow} + \sqrt{\frac{2}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{3}{2}-\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{2}{3}}\Phi_{210\downarrow} \\ \Psi_{21\frac{3}{2}-\frac{3}{2}} &= \Phi_{21-1\downarrow} \end{align}$$

And the term $^2P_{1/2}$ (i.e. with $s=\frac{1}{2},l=1,j=\frac{1}{2}$) has $2$ configurations ($m_j=+\frac{1}{2},-\frac{1}{2}$): $$\begin{align} \Psi_{21\frac{1}{2}+\frac{1}{2}} &= \sqrt{\frac{2}{3}}\Phi_{21+1\downarrow} - \sqrt{\frac{1}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{1}{2}-\frac{1}{2}} &= - \sqrt{\frac{2}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{1}{3}}\Phi_{210\downarrow} \end{align}$$

You can write the states $\Psi_{nljm_j}$ (which are eigenvectors of $J^2$ and $J_z$) as certain superpositions of the states $\Phi_{nlm_lm_s}$ (which are eigenvectors of $L_z$ and $S_z$). But unfortunately this is quite an elaborate task, and usually text books on quantum mechanics take at least one whole chapter to explain this. You have already figured out the number of configurations for the terms. The final result (with the coefficients taken from the table of Clebsch-Gordan coefficients, section 2.3) is this.

The term $^2P_{3/2}$ (i.e. with $l=1,j=\frac{3}{2}$) has $4$ configurations ($m_j=+\frac{3}{2},+\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$): $$\begin{align} \Psi_{21\frac{3}{2}+\frac{3}{2}} &= \Phi_{21+1\uparrow} \\ \Psi_{21\frac{3}{2}+\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21+1\downarrow} + \sqrt{\frac{2}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{3}{2}-\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{2}{3}}\Phi_{210\downarrow} \\ \Psi_{21\frac{3}{2}-\frac{3}{2}} &= \Phi_{21-1\downarrow} \end{align}$$

And the term $^2P_{1/2}$ (i.e. with $l=1,j=\frac{1}{2}$) has $2$ configurations ($m_j=+\frac{1}{2},-\frac{1}{2}$): $$\begin{align} \Psi_{21\frac{1}{2}+\frac{1}{2}} &= \sqrt{\frac{2}{3}}\Phi_{21+1\downarrow} - \sqrt{\frac{1}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{1}{2}-\frac{1}{2}} &= - \sqrt{\frac{2}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{1}{3}}\Phi_{210\downarrow} \end{align}$$

You can write the states $\Psi_{nljm_j}$ (which are eigenvectors of $J^2$ and $J_z$) as certain superpositions of the states $\Phi_{nlm_lm_s}$ (which are eigenvectors of $L_z$ and $S_z$). But unfortunately this angular momentum coupling is quite an elaborate task, and usually text books on quantum mechanics take at least one whole chapter to explain it. You have already figured out the number of configurations for the terms. The final result (with the coefficients taken from the table of Clebsch-Gordan coefficients, section 2.3) is this.

The term $^2P_{3/2}$ (i.e. with $l=1,j=\frac{3}{2}$) has $4$ configurations ($m_j=+\frac{3}{2},+\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$): $$\begin{align} \Psi_{21\frac{3}{2}+\frac{3}{2}} &= \Phi_{21+1\uparrow} \\ \Psi_{21\frac{3}{2}+\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21+1\downarrow} + \sqrt{\frac{2}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{3}{2}-\frac{1}{2}} &= \sqrt{\frac{1}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{2}{3}}\Phi_{210\downarrow} \\ \Psi_{21\frac{3}{2}-\frac{3}{2}} &= \Phi_{21-1\downarrow} \end{align}$$

And the term $^2P_{1/2}$ (i.e. with $l=1,j=\frac{1}{2}$) has $2$ configurations ($m_j=+\frac{1}{2},-\frac{1}{2}$): $$\begin{align} \Psi_{21\frac{1}{2}+\frac{1}{2}} &= \sqrt{\frac{2}{3}}\Phi_{21+1\downarrow} - \sqrt{\frac{1}{3}}\Phi_{210\uparrow} \\ \Psi_{21\frac{1}{2}-\frac{1}{2}} &= - \sqrt{\frac{2}{3}}\Phi_{21-11\uparrow} + \sqrt{\frac{1}{3}}\Phi_{210\downarrow} \end{align}$$