Consider the Reed Solomon code with parameters $[n,k,d=n-k+1]$ over the field $\mathbb{F}_q.$ Take $q=2^m,$ and note that the usual choice is to choose $n=q-1,$ for $q-$ary symbols. Therefore the binary length of this code is $N=nm=(2^m-1)m$ while its binary "input length" is $K=km.$
This code is MDS, its full weight distribution (over $\mathbb{F}_q$) is known but we only need the number of codewords of minimum weight which is $$A_d=\binom{n}{d}(q-1).$$
Now choose, say, $k=2n/3$ (one can assume $3|n$ ) which gives $d=n/3$.
We use the codewords of this code to define a binary map $$f:\{0,1\}^K\rightarrow \{0,1\}^N.$$
If $x \in Im(f)$ i.e., $x$ is the binary image of a Reed Solomon codeword, there are $$A_d=A_{n/3}=\binom{n}{n/3} (q-1)\sim 2^{nH(1/3)+o(n)+m}
\stackrel{(a)}{\sim} 2^{0.92 n+m} =$$
$$=\exp_2\left\{(nm)\left[0.92/m+o(1/m)+(1/n)\right] \right\}\sim2^{0.92 (3km/2m)}=$$
$$=2^{0.92 (3K/2m)}=2^{1.38 K/m}\leq 2^{K(1-\delta)},\quad \mathrm{for~m~large~enough},$$
codewords within the $K/2$ sphere around it (to justify (a), see the answer here for the binomial approximation).
The covering radius of the Reed Solomon codes is known to be $d-1,$ and if $x$ is a point at maximal distance from a codeword, it cannot possibly have more neighbours within the $K/2$ sphere than the number of neighbours of a codeword. This can be verified by using the triangle inequality. This argument also applies to any non-codeword $x,$ thus establishing the claim that PRNG functions satisfying the requirements of the OP exist.
Note that reed Solomon codewords can be constructed by matrix multiplication or polynomial evaluation, thus with $\mathrm{poly}(n)=\mathrm{poly}(K)$ complexity.