You start with:
$$ \psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} \tag{1} $$
Since $\psi_{sp^3}$ is normalised we know that:
$$ \langle \psi_{sp^3} | \psi_{sp^3} \rangle = 1 $$
and if we use equation (1) to substitute for $\psi_{sp^3}$ we get:
$$ \langle c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} | c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y}+ c_4\psi_{2p_{z}} \rangle = 1 $$
and expanding gives:
$$ \langle c_1\psi_{2s} | c_1\psi_{2s} \rangle + \langle c_1\psi_{2s} | c_2\psi_{2p_{x}} \rangle \, + ... + \, \langle c_4\psi_{2p_{z}} | c_3\psi_{2p_{y}} \rangle + \langle c_4\psi_{2p_{z}} | c_4\psi_{2p_{z}} \rangle = 1 $$
I've skipped the middle twelve terms because, well, life's too short and it should be obvious what they are. We can take the constants $c$ out of the brackets to get:
$$ c_1^2\langle \psi_{2s}|\psi_{2s} \rangle + c_1c_2\langle\psi_{2s}|\psi_{2p_{x}} \rangle \, + ... + \, c_4c_3\langle\psi_{2p_{z}} | \psi_{2p_{y}} \rangle + c_4^2\langle \psi_{2p_{z}} | \psi_{2p_{z}} \rangle = 1 $$
But the $2s$ and $2p$ orbitals are all orthonormal which means that $\langle\psi_i|\psi_j\rangle = \delta^i_j$, so our equation becomes:
$$ c_1^2 + c_2^2 + c_3^2 + c_4^2 = 1 $$
Which I think is what you were asking.
Response to comment:
The coefficients don't rotate anything, they are just numbers. What the diagram shows is that if you take two vectors:
then you can get a vector $\psi$ in any direction you want using:
$$ \psi = c_1\psi_1 + c_2 \psi_2 $$
with the normalisation condition $c_1^2 + c_2^2 = 1$. For example:
$$ \psi = \frac{1}{\sqrt{2}}\psi_1 + \frac{1}{\sqrt{2}}\psi_2 $$
is at 45ยบ. More generally:
$$ \psi = \sin(\theta)\psi_1 + \cos(\theta)\psi_2 $$
gives you a vector at an angle $\theta$ to the horizontal, and it's automatically normalised since $\sin^2 + \cos^2 = 1$.
The diagram is just showing this applies to the $2p$ orbitals. Combining $2p_x$ and $2p_y$ in different proportions gives an orbital at different angles.
As a proxy for an atom's reluctance to form chemical bonds, we can use the energy required to lift one of its outer electrons to the next higher orbital, pretending as usual that the "orbital" concept is still at least approximately valid in a multi-electron atom. Then the question is why this energy gap tends to be significantly larger after filling a p-shell ($\ell=1$) than it is after filling s- or d- or f-shells ($\ell=0,2,3$).
As usual, let $n$ and $\ell$ denote the radial and angular quantum numbers, respectively. Let's accept that orbitals are filled in order of increasing $n+\ell$, and then in order of increasing $n$ whenever the first rule is neutral. (Reality is a little more complicated, but these rules mostly work pretty well.) The following list shows the filling order defined by these two rules, along with the number of nodes in the orbital wavefunction's radial factor:
\begin{align}
n+\ell &\hskip1cm (n,\ell) &\hskip1cm \text{radial nodes} & \\
\hline
1 &\hskip1cm (1,0) &\hskip1cm 0 &\hskip1cm \text{helium}\\
2 &\hskip1cm (2,0) &\hskip1cm 1 \\
3 &\hskip1cm (2,1) &\hskip1cm 0 &\hskip1cm \text{neon}\\
3 &\hskip1cm (3,0) &\hskip1cm 2 \\
4 &\hskip1cm (3,1) &\hskip1cm 1 &\hskip1cm \text{argon} \\
4 &\hskip1cm (4,0) &\hskip1cm 3 \\
5 &\hskip1cm (3,2) &\hskip1cm 0 \\
5 &\hskip1cm (4,1) &\hskip1cm 2 &\hskip1cm \text{krypton} \\
5 &\hskip1cm (5,0) &\hskip1cm 4 \\
6 &\hskip1cm (4,2) &\hskip1cm 1 \\
6 &\hskip1cm (5,1) &\hskip1cm 3 &\hskip1cm \text{xenon} \\
6 &\hskip1cm (6,0) &\hskip1cm 5 \\
7 &\hskip1cm (4,3) &\hskip1cm 0 \\
7 &\hskip1cm (5,2) &\hskip1cm 2 \\
7 &\hskip1cm (6,1) &\hskip1cm 4 &\hskip1cm \text{radon} \\
7 &\hskip1cm (7,0) &\hskip1cm 6 \\
8 &\hskip1cm (5,3) &\hskip1cm 1 \\
\end{align}
In every noble-gas case, and only in these cases, the next available level is an s-orbital ($\ell=0$), which has more radial nodes than the levels immediately before or after it in the sequence. (This pattern works for helium, too, even though helium doesn't have a filled p-shell.) If we could understand intuitively why energy gap between the ground state and the first excited state is relatively large when the first excited state is an s-orbital, then we would have at least a partial answer to the question.
The order in which the shells are filled (the sequence shown above) indicates that radial nodes are more costly than "azimuthal" nodes, because for a given $n+\ell$, the cases with smaller $\ell$ (fewer "azimuthal" nodes) have higher energy. Accepting this trend as an axiom, we can focus our intuition on the radial part.
Intuitively, if we think of the radial nodes as "no-fly zones" for that electron, then having a larger number of radial nodes may correspond to having less freedom to re-arrange the multi-electron system to minimize the energy in the presence of electron-electron interactions. Here's the key idea: by analogy with a construction-induced traffic jam, the impact of each additional node (each additional no-fly zone) could be an increasing function of the number of nodes already present. This picture suggests that lifting an electron from a filled p-shell up to the next available s-shell should be more costly than, say, lifting an electron from a filled d-shell up to the next available p-shell, because the former lift requires increasing an already-larger number of radial nodes. Using the magnitude of the energy-gap as a proxy for an atom's reluctance to form bonds, this suggests that the noble gases should be less reactive (relatively), at least among atoms with the same value of $n+\ell$.
The key idea is that the energy-cost of each additional radial node is an increasing function of the number of radial nodes already present. This seems to be consistent with the information I've seen, but I have no real justification for anticipating it except for the dubious traffic-jam analogy. Even if the intuition is correct, it isn't quantitative enough to predict just how noble the noble gases are. It is suggestive at best, but at least it doesn't rely entirely on a computer. That's why I thought it was worth posting.
Best Answer
In general, atoms need not be spherically symmetric.
The source you've given is flat-out wrong. The wavefunction it mentions, $\varphi=\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$, is in no way spherically symmetric. This is easy to check: the wavefunction for the $2p_z$ orbital is $\psi_{2p_z}(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:z \:e^{-r/2a_{0}}$ (and similarly for $2p_x$ and $2p_y$), so the wavefunction of the combination is $$\varphi(\mathbf r)=\frac {1}{\sqrt {32\pi a_0^5}}\:\frac{x+y+z}{\sqrt 3} \:e^{-r/2a_{0}},$$ i.e., a $2p$ orbital oriented along the $(\hat{x}+\hat y+\hat z)/\sqrt3$ axis.
This is an elementary fact and it can be verified at the level of an undergraduate text in quantum mechanics (and it was also obviously wrong in the 1960s). It is extremely alarming to see it published in an otherwise-reputable journal.
On the other hand, there are some states of the hydrogen atom in the $2p$ shell which are spherically symmetric, if you allow for mixed states, i.e., a classical probabilistic mixture $\rho$ of hydrogen atoms prepared in the $2p_x$, $2p_y$ and $2p_z$ states with equal probabilities. It is important to emphasize that it is essential that the mixture be incoherent (i.e. classical and probabilistic, as opposed to a quantum superposition) for the state to be spherically symmetric.
As a general rule, if all you know is that you have "hydrogen in the $2p$ shell", then you do not have sufficient information to know whether it is in a spherically-symmetric or an anisotropic state. If that's all the information available, the initial presumption is to take a mixed state, but the next step is to look at how the state was prepared:
It is extremely tempting to think (as discussed previously e.g. here, here and here, and links therein) that the spherical symmetry of the dynamics (of the nucleus-electron interactions) must imply spherical symmetry of the solutions, but this is obviously wrong $-$ to start with, it would apply equally well to the classical problem! The spherical symmetry implies that, for any anisotropic solution, there exist other, equivalent solutions with complementary anisotropies, but that's it.
The hydrogen case is a bit special because the $2p$ shell is an excited state, and the ground state is symmetric. So, in that regard, it is valid to ask: what about the ground states of, say, atomic boron? If all you know is that you have atomic boron in gas phase in its ground state, then indeed you expect a spherically-symmetric mixed state, but this can still be polarized to align all the atoms into the same orientation.
As a short quip: atoms can have nontrivial shapes, but the fact that we don't know which way those shapes are oriented does not make them spherically symmetric.
So, given an atom (perhaps in a fixed excited state), what determines its shape? In short: its term symbol, which tells us its angular momentum characteristics, or, in other words, how it interacts with rotations.
A good comprehensive source for term symbols of excited states is the Levels section of the NIST ASD.