We proceed by induction on \(k\text{.}\) In the case \(k=1\text{,}\) we have that \(\mathcal{F}\) is a collection of at least \(p\) distinct single-element sets, and so any \(p\) of them form a sunflower with core \(\emptyset\text{.}\)

Now, suppose that \(k\geq2\text{.}\) Let \(\mathcal{A} = \{A_1,\dots,A_t\}\) be a maximal collection of pairwise disjoint elements of \(\mathcal{F}\text{.}\) If \(t\geq p\text{,}\) then we are done, as the sets in \(\mathcal{A}\) form a sunflower with core \(\emptyset\text{.}\) So, we assume that \(t\leq p-1\text{.}\) Define

\begin{equation*}
X:=\bigcup_{i=1}^tA_i\text{.}
\end{equation*}

Since every set in \(\mathcal{F}\) has cardinality \(k\) and \(t\leq p-1\text{,}\) we have that \(|X|\leq (p-1)k\text{.}\) By maximality of \(\mathcal{A}\text{,}\) every element of \(\mathcal{F}\) intersects \(X\text{.}\) Therefore, by the Pigeonhole Principle, there is a point \(x\in X\) which is contained in at least

\begin{equation*}
\frac{|\mathcal{F}|}{|X|} > \frac{k!(p-1)^k}{(p-1)k}=(k-1)!(p-1)^{k-1}
\end{equation*}

sets of \(\mathcal{F}\text{.}\) Consider the system

\begin{equation*}
\mathcal{F}':= \{F\setminus \{x\}: F\in \mathcal{F}\text{ and } x\in F\}\text{.}
\end{equation*}

Then \(\mathcal{F}'\) is a collection of more than \((k-1)!(p-1)^{k-1}\) sets of cardinality \(k-1\text{;}\) therefore, it has a sunflower with \(p\) petals by the induction hypothesis. Adding the element \(x\) back into each set of this sunflower gives us a sunflower with \(p\) petals in \(\mathcal{F}\text{.}\)