= Solution
View the oriented square lattice as $(x,n)$ with $x+n$ even, and arrows $(x,n)\to(x\pm1,n+1)$. Each <graph vertex> has two incoming bonds. Start with independent <oriented bond percolation> of parameter $p$, and declare a site open when at least one of its incoming bonds is open. Its <probability> of being open is
$$
r=1-(1-p)^2.
$$
Incoming bond sets for different heads are disjoint, so these site indicators are independent: they are precisely an <oriented site percolation> configuration with parameter $r$.
Every <graph vertex> after the first on an open oriented bond path is open as a site, because its incoming path bond is open. The root's two incoming bonds belong to earlier time layers and are independent of all bonds used by a forward path from the root. Thus the event that the root site is open, of <probability> $r$, is independent of bond survival from that root. On their intersection the entire infinite bond path is also an infinite open site path. Hence, writing $\theta_b,\theta_s$ for the two survival <probabilities>,
$$
\theta_s(1-(1-p)^2)\ge\bigl(1-(1-p)^2\bigr)\theta_b(p).
$$
This is the <incoming-edge coupling of site and bond percolation>. If site survival is defined conditional on an open root, the extra factor is omitted; the critical <probability> is the same. If the model is drawn only in a time half-plane, one can independently sample the root's incoming bonds as auxiliary variables.
For every $p>p_c(\mathrm{bond})$, bond survival is positive by monotonicity and the definition of the critical <probability>. Therefore $p_c(\mathrm{site})\le1-(1-p)^2$. Let $p$ decrease to the bond critical <probability> and use continuity of this polynomial, obtaining
$$
\boxed{p_c(\mathrm{site})\le1-(1-p_c(\mathrm{bond}))^2.}
$$
If the bond critical <probability> equals one, the inequality is trivial; no assertion about survival exactly at criticality is used.
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