This is expanding on my earlier comment. Though it's not a full answer, it does say something about the case $p > 1$.

Given sets $S_1,\ldots,S_k$ which satisfy the condition for a given $p \geq 1$. Let $T_1,\ldots,T_\ell$ list all the sets that can be obtained by deleting exactly $p-1$ elements from one of the sets $S_i$. Thus $$\ell = \binom{|S_1|}{p-1} + \cdots + \binom{|S_k|}{p-1}$$ and observe that $\{T_1,\ldots,T_\ell\}$ is a Sperner family. It follows from Sperner's Theorem that $$\ell \leq \binom{n}{\lfloor n/2 \rfloor},$$ which gives a lower bound on the cardinality $n$ of the union $S_1 \cup \cdots \cup S_k = T_1 \cup \cdots \cup T_\ell$ in terms of $\ell$.

When $p = 1$, this is optimal since the collection of all subsets of $\{1,\ldots,n\}$ with size exactly $\lfloor n/2 \rfloor$ reaches the exact bound given above. When $p > 1$, we get something more crude by observing that we obviously have $|S_i| \geq p$ for every $i$ and thus $$kp \leq \binom{n}{\lfloor n/2 \rfloor}.$$

Finally, here is a simple construction of a large family of subsets of $\{1,\ldots,n\}$ that has the property with $p = 2$. For $z = 0,\ldots,n-1$, consider $$\mathcal{S}_z = \{S \subseteq \{1,\ldots,n\} : |S| = \lfloor n/2 \rfloor, {\textstyle\sum_{x \in S} x \equiv z \pmod{n}}\}.$$ All of these families have the given property for $p = 2$. Since $$|\mathcal{S}_0| + \cdots + |\mathcal{S}_{n-1}| = \binom{n}{\lfloor n/2 \rfloor},$$ at least one of these families satisfies $$|\mathcal{S}_z| \geq \frac{1}{n}\binom{n}{\lfloor n/2 \rfloor} = \binom{n-1}{\lceil (n-1)/2 \rceil}.$$

To summarize, if $\nu(k,p)$ denotes the minimum size of the union of a family of $k$ sets with the given property for a given $p$, then:

- $\nu(k,1) = \min\left\{n : k \leq \binom{n}{\lfloor n/2 \rfloor}\right\}$
- $\nu(k,p) \geq \min\left\{n : kp \leq \binom{n}{\lfloor n/2 \rfloor}\right\}$
- $\nu(k,2) \leq \min\left\{n+1 : k \leq \binom{n}{\lfloor n/2 \rfloor}\right\}$