K = 01FE01FE01FE01FE P = 771DF32699DF2F2A C = 88E20CD96620D0D5
K = 01FE01FE01FE01FE P = 668493768A3D5786 C = 997B6C8975C2A879
K = 01FE01FE01FE01FE P = 6F52785BC50C62BE C = 90AD87A43AF39D41
K = 01FE01FE01FE01FE P = DFFE8B7E72E5CFED C = 200174818D1A3012
will do. I found that by a randomized search after about $2^{34}$ DES operations, which tends to confirm the order of magnitude of the $2^{32}$ stated in the question.
Explanation:
For $i\in[1,16]$, note $L_{i-1}$ and $R_{i-1}$ the 32-bit contents of the $L$ and $R$ registers before encryption round $i$; note $R_{16}$ and $L_{16}$ the output of the last round after a swap (even if that swap is undone by an extra final swap, or equivalently is not done); note $K_i$ the 48-bit sub-key for round $i$. We have $L_i=R_{i-1}$ and $R_i=L_{i-1}\oplus F(E(R_{i-1})\oplus K_i)$, where $E$ is the expansion, and $F$ represents the S-boxes and permutation $P$ combined.
For the four keys $K$ 01FE01FE01FE01FE
, FE01FE01FE01FE01
, 1FE01FE00EF10EF1
, E01FE01FF10EF10E
, it holds that $K_i=\overline{K_{17-i}}$.
If it happens that $L_{8}=\overline{R_{8}}$, using that $E(\overline X)\oplus \overline{K_i}=\overline{E(X)}\oplus \overline{K_i}=E(X)\oplus K_i$, it follows that the input of function $F$ (S-boxes and permutation $P$) at rounds 8 and 9 are the same, thus their output are the same. By induction we get that the input of $F$ at round $i$ and $17-i$ are the same, and $L_i=\overline{R_{16-i}}$. Thus $L_{16}=\overline{R_0}$ and $R_{16}=\overline{L_0}$, thus $E_K(P)=\overline P$ for the $P$ corresponding to $L_0$ and $R_0$.
We could efficiently construct $2^{32}$ plaintexts $P$ with that property by performing the last 8 rounds of decryption and final permutation of DES, starting from the $2^{32}$ values of $L_{8}=\overline{R_{8}}$.
Unsettled: does there exist a DES key $K$ with more than $2^{32}$ anti-fixed points?