Mathematical form of four hybrid orbitals The mathematical form of the four $\ce{sp^3}$ hybrid orbitals are given below.
  \begin{align}
\ce{
\tag{a} sp^3 &amp;= 1/2s + 1/2p_x + 1/2p_y + 1/2p_z\\
\tag{b} sp^3 &amp;= 1/2s + 1/2p_x - 1/2p_y - 1/2p_z\\
\tag{c} sp^3 &amp;= 1/2s - 1/2p_x + 1/2p_y - 1/2p_z\\
\tag{d} sp^3 &amp;= 1/2s - 1/2p_x - 1/2p_y - 1/2p_z\\
}\end{align}
  For each hybrid orbital the number in front of the $\ce{s}$ and three $\ce{p}$ functions, called a coefficient, describes the contribution and relative ratio of each canonical orbital to the hybrid wave function. Add up the coefficients and prove to that these orbitals are $\ce{s^1p^3}$.


Yes, I know that the grammar is terrible on that last sentence, that is how it is written. So, I know that for an $\ce{sp^3}$ orbital it is one part $\ce{s}$ orbital and three parts $\ce{p}$ orbital. However, I have no idea what this question wants me to do. Adding up the coefficients gives me $2$, $0$, $0$, and $-1$, but I have no idea if that is how it wants me to add it up (you could ignore the sign and get $2$ every time, and show that $1/2$ is a quarter of $2$, thus proving that each component contributes a quarter to the hybridized orbital, but this seems too simple and incorrect).
If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination

$$\psi = \sum_n c_n \phi_n$$

then loosely speaking, the "amount" of $\phi_n$ in $\psi$ is given not by $c_n$, but rather by $\left|c_n\right|^2$. (See, for example, Griffiths, Introduction to Quantum Mechanics, 2nd ed., Section 3.4.)

Therefore, summing the four values of $c_n$ for each AO has no physical meaning.

Instead let's take wavefunction (b) for example. The coefficient in front of the s orbital is $1/2$, so the "amount" of s-orbital character is simply $\left|1/2\right|^2 = 1/4$. The coefficient in front of the $\mathrm{p}_y$ orbital is $-1/2$, so there is $\left|-1/2\right|^2 = 1/4$ $\mathrm{p}_y$-character.

Armed with this knowledge, it's incredibly easy to see that each orbital has exactly $1/4$ s-character and a total of $3/4$ p-character ($1/4$ from each p orbital), because none of the $\pm$ signs matter (the minus signs are removed by the absolute value).