It isn't true. Choose $n,a,k$ such that $n+a-k>0$, $n+a-2k<0$ and $k>1$. The sum has only one nonzero term which is not equal to the right side.
Using my Maple package BinomSums it is easy to compute the generating function of the left-hand side of your identity (let's call it $u_{n,m,p,n',m',p'}$):
$$\sum_{n,m,p,n',m',p' \geq 0} u_{n,m,p,n',m',p'} x_1^n y_1^m z_1^p x_2^{n'} y_2^{m'} z_2^{p'} = -\frac{(x_2-1)(x_1-1)}{(y_2z_2+x_2-1)(y_1x_2-y_2x_2-y_1-x_2+1)(y_1x_1-y_2x_1+y_2+x_1-1)(y_1z_1+x_1-1)}$$
Thus a rational generating function usually gives all we need about a sequence.
Here is the code:
S := Sum(Sum(Binomial(n1+p1,i1)*Binomial(n2+p2,i2)*Multinomial([i1-p1,m2-i2])*Multinomial([i2-p2,m1-i1]), i1=0..infinity), i2=0..infinity);
vars := [n1,m1,p1,n2,m2,p2]:
gfunS := foldl(Sum, S*mul((t||v)^v, v in vars), seq(v=0..infinity, v in vars));
BinomSums[sumtores](gfunS, u);
Best Answer
In terms of hypergeometric series, the sum is $_3F_2(-n, 1+n, 1/2;1,3/2;1)$ and the identity is a special case of Saalschütz's theorem (also called the Pfaff-Saalschütz theorem), one of the standard hypergeometric series identities.
A more general identity, also a special case of Saalschütz's theorem, is $$\sum_{k=0}^n (-1)^k\frac{a}{a+k}\binom{n+k+b}{n-k}\binom{2k+b}{k} = \binom{n+b-a}{n}\biggm/\binom{n+a}{n}.$$ The O.P.'s identity is the case $a=1/2, b=0$.