Forget all this Hermite and Laguerre stuff. The fact is that any family of polynomials with all different degrees is linearly free. Hence any family of polynomials with degrees $0$, $1$, $\ldots$, $n$ is a basis of the vector space of polynomials of degree at most $n$ (the space you denote by $\mathbb{P}_n$).
A relatively more sophisticated way of saying the same thing is that any triangular matrix with no zero on its diagonal is invertible.
There is, in general, no general mnemonic for the $K_n $. Those constants are fixed by convention and such conventions are different between the big families of classical OPs, which reflects the different situations they're used in. No instructor should expect you to memorize them. For problem solving, or as a working research phycisist, the usual is to have a copy of Abramowitz & Stegun, or simply the DLMF, to refer to for the precise form of the normalization factors in such formulae.
Edit in response to comment:
To be clear, this is not doable. The conventions are chosen independently for the different families and respond to different pressures; there is no overarching scheme. What Szegö is doing in your quote is reducing the Rodrigues formula to the form $\text{const}\times P_n^{(\alpha,\beta)}$ and then fixing the normalization using the fact that
$$
P_n^{(\alpha,\beta)}(1)=\begin{pmatrix}n+\alpha\\n\end{pmatrix},
$$
which is equally arbitrary.
Have a look at the table with the definitions of the different families in the DLMF (the notation for which is here) and at the table of special values. Some are normalized to a constant norm (e.g. Chebyshev), some are normalized to constant leading coefficient (e.g. Hermite $\rm{He}_n$, Chebyshev), some are normalized to a constant value at $x=0$ or $x=1$ (e.g. Legendre). Others are normalized due to other considerations (like the Hermite $H_n$, which are normalized so $H_n(x)e^{-x^2/2}$ is an eigenfunction of the Fourier transform).
What you want is impossible. If you want the normalization constants at your fingertips, memorize them the hard way, or have the DLMF or the NIST Handbook or Abramowitz and Stegun at your fingertips.
Best Answer
Well, the best way I can think of to introduce a real physical problem where Laguerre ànd Legendre (even of the associated form!) differential equations play a role, is the (exact, that is, without taking spin into consideration) solution of the combined angular momentum / magnetic moment / energy eigenvalue-problem for the hydrogen atom.
I cannot tell the whole story here, but I can sketch the process.
When solving the Schrodinger equation involved, and transforming it to spherical coordinates, all applicable quantum numbers roll into your lap and on route you will encounter the Laguerre and associated Legendre differential equations.
Now for the approach of solving these equations. Let's have a look at the Laguerre equation first.
$$xy'' + (1+\alpha-x)y' + ny = 0$$
If you try $y = \sum_{j=0}^{\infty}a_jx^j$, you will find that $$\frac{a_{j+1}}{a_j} = -\frac{n-j}{(j+1)(j+1+\alpha)}$$ So, when moving from one coefficient to the next, the sign changes, it is divided by $j+1$, multiplied by $n-j$ and divided by $j+1+\alpha$. When j reaches n, the coefficients vanish. That means that we will not be far off if we try $$a_j = \frac{(-1)^j}{j!}\binom{n+\alpha}{n-j}$$ In fact, this is exactly what we need.
Now, which function will this be? The factor $\frac{(-1)^j}{j!}$ points at $e^{-x}$. The factor $\binom{n+\alpha}{n-j}$ points at the $j^{th}$ derivative of $x^{n+\alpha}$, and we saw above that the series ends for j=n. But when we try $[e^{-x}x^{n+\alpha}]^{(n)}$, firstly we have to compensate for all the $e^{-x}$ factors by multiplying the result by $e^x$ , and secondly we would get terms for the running index j that have a factor $x^{n + \alpha - (n-j)} = x^{j + \alpha}$ and we would like to have $x^j$. So we have to compensate at the end here too, by multiplying by $x^{-\alpha}$. This reasoning leads to trying $$y = e^xx^{-\alpha}[e^{-x}x^{n+\alpha}]^{(n)}$$ Where $f^{(n)}$ stands for the $n^{th}$ derivative of $f$. And funnily enough, it's spot on again.
Over to the Legendre equation, first the ordinary one.
$$[(1-x^2)y']' + n(n+1)y = 0$$
Trying $y = \sum_{j=0}^{\infty}a_jx^j$ again, we find that $$\frac{a_{j+2}}{a_j} = \frac{j(j+1) - n(n+1)}{(j+1)(j+2)}$$ As can be seen in one of my other posts (About the Legendre differential equation) there are solutions that are polynomials ($P_n$) and solutions that are represented by an unending series expansion ($Q_n$). The $P_n$ are either odd-powered or even powered polynomials, depending on $n$ which is also the degree of $P_n$.
The $Q_n$ contain a factor $ln(\frac{1+x}{1-x})$, see (About the Legendre differential equation) again.
Finally, the associated Legendre equation. $$[(1-x^2)y']' + \{l(l+1) - \frac{m^2}{1-x^2}\}y = 0$$ It is natural to try $y = (1-x^2)^{\alpha}P_l$. It is easily checked that we get the desired term $\frac{m^2}{1-x^2}y$ if we take $\alpha = m/2$. Further analysis shows that if we wish to arrive at all the correct terms, we have to use $P_l^{(m)}$ instead of $P_l$. Note that I still use the notation $f^{(m)}$ as denoting the $m^{th}$ derivative of $f$. The notation I use for the solutions of the associated Legendre equation is $$P_{l,m} = (1-x^2)^{m/2}P_l^{(m)}$$
I hope this helps :-).