Without the inequality on the $i, j, k$, the sum on the left would be $\binom{3p}{p}$. This is an immediate consequence of the Chu-Vandermonde convolution identity

$$\sum_{i+j = k} \binom{x}{i}\binom{y}{j} = \binom{x+y}{k}$$

which is treated in many places, for example in Concrete Mathematics by Graham, Knuth, and Patashnik.

By an inclusion-exclusion argument, the adjustment one must make to account for the inequality in the sum is to subtract the sum where at least one of the $i, j, k$ is $0$, then add back in the sum where at least two of them are $0$. For example, the sum over all cases where $k = 0$ is

$$\sum_{i+j=p} \binom{p}{i}\binom{p}{j} = \binom{2p}{p}$$

where again we use Chu-Vandermonde. This leads to the answer you want.