<s>[INST] Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series. 
 [/INST] Ok, here is an approach with generating functions. Let
$$
   g_1(z) = \sum_{s=0}^\infty \binom{p+s}{s} z^s = \frac{1}{\left(1-z\right)^{p+1}}
$$
$$
   g_2(z) = \sum_{s=0}^\infty \binom{2p+m}{s} z^s = \left(1+z\right)^{m+2p}
$$
Now
$$ \begin{eqnarray}
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} &=& \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{m-2s} = [z]^m g_1(z^2) g_2(z) = [z]^m \frac{\left(1+z\right)^{m+2p}}{(1-z^2)^{p+1}} \\
   &=& [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}}
 \end{eqnarray} 
$$
Here is a verification:
In[27]:= With[{p = 5, 
  m = 7}, {SeriesCoefficient[(1 + z)^(m + 2 p)/(1 - z^2)^(
   p + 1), {z, 0, m}], 
  Sum[Binomial[p + s, s] Binomial[2 p + m, 2 p + 2 s], {s, 
    0, \[Infinity]}]}]

Out[27]= {71808, 71808}

Let's continue:
$$ \begin{eqnarray}
    [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}} &=& \sum_{s=0}^\infty \binom{p+m-1}{m-s} \binom{p+s}{s} =  \sum_{s=0}^\infty \binom{p+m-1}{p+s-1} \binom{p+s}{s}\\ &=& \sum_{s=0}^\infty \frac{(p+s) (m+p-1)!}{p! s! (m-s)!} = 
        \sum_{s=0}^\infty \frac{p (m+p-1)!}{p! s! (m-s)!} + \sum_{s=0}^\infty \frac{s   (m+p-1)!}{p! s! (m-s)!} \\
   &=& \binom{m+p-1}{m} \left( \sum_{s=0}^\infty  \binom{m}{s} + \sum_{s=0}^\infty \frac{s}{p} \binom{m}{s}  \right) \\ 
    &=& \binom{m+p-1}{m}  \left( 2^m + 2^{m-1} \frac{m}{p} \right) 
\end{eqnarray}
$$
</s>