ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$.
Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that:
$$ I = (I,x) \cap (I,F_1)$$
If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$.
Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies
$$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$
$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired.
REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question.
So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p).
It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show:
$$I \cap J = P + I'J' $$
Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that this is even true for $I,J$ generated by parts of a fixed regular sequence).
Best Answer
As Sandor pointed out, a necessary condition is that the prime ideal $P$ is a complete intersection. Here is a proof that it is also sufficient. It will suffice to prove the following:
Claim: Let $(R,m)$ be a Noetherian local ring and $x\in m$ a regular element on $R$. If $R/(x)$ is a domain, then so is $R$.
Proof: Suppose $ab=0$ in $R$. Then modulo $x$, one of them say $a$, must be $0$. So $a=xa_1$, thus $x(a_1b)=0$. As $x$ is regular, $a_1b=0$, and continuing in this fashion one of $a,b$ must be divisible by arbitrary high power of $x$, so it must be equal to $0$.
As for an example which is not a part of a regular s.o.p, take something like $P=(x^2+y^2+z^2, u^2+v^2+w^2)$ in $\mathbb C[[x,y,z,u,v,w]]$.