Upper bounds for a binomial coefficient
The argument looks correct. Also notice that you can get a better (but still loose) upper bound as follows:
$$
\binom{k}{p-1} \le \sum_{i=0}^{k} \binom{k}{i} = 2^k
$$
Where the equality $\sum_{i=0}^{k} \binom{k}{i} = 2^k$ follows from the fact that the summation on the left is counting the number of possible subsets of a set with $k$ elements, grouped by cardinality: the $i$-th term of the sum (for $i=0, \dots, k$) is the number of subsets with exactly $i$ elements.
Also, using Stirling's approximation and assuming that $k$ is even (this is just for convenience, if it's not you can consider $k+1$ instead to get the same asymptotic bound):
$$
\binom{k}{p-1}  \le \binom{k}{k/2} = O\left( \frac{ (k/e)^k \sqrt{k} }{ k (k/2e)^k } \right) =
O\left(\frac{2^k}{\sqrt{k}} \right).
$$