Conditional $k,n$ secret sharing
This can be trivially achieved by sharing twice.
First share your secret $S$ using a $2$-out-of-$2$ secret sharing scheme into $S_1$ and $S_2$. (E.g. choose $S_1$ at random and set $S_2:=S\oplus S_1$.)

In the next step you now share $S_2$ using a $(k-1)$-out-of-$(n-1)$ secret sharing scheme into $S_{2,1},\dots,S_{2,n-1}$.

Give $S_1$ to $P_1$ and $S_{2,i}$ to $P_{i+1}$ for $1\leq i &lt;n$.

The secrecy of the $2$-out-of-$2$ secret sharing scheme guarantees that both shares $S_1$,$S_2$ are required. $S_1$ is only known to $P_1$, and $S_2$ can only be recovered by at least $k-1$ of the other parties collaborating.