It's 1 to 1 in a neighborhood of 0, and so the identity is also. But as you say, it can't be true over all p-adic algebraic integers, even when restricted to positive valuation. Try to compute the radius of convergence of the log function. Note that all of this has an analogue over the complex numbers

Yes. Look up Gauss lattice basis reduction, and apply it to $\{(x\pmod{p^k} , 1) , (p^k,0)\}$.

Did you have luck with this?

If $E/F$ is a finite field extension (not characteristic 2) and $\pi$ is a generic, principal, non-Steinberg representation, that is not the normalised induction of a pair of characters of $E^\times$ which are distinct, and both trivial on $F^\times$, then I think that $(\pi, 1)_{B(F)}=0$. Are you sure that over local fields the hom-space is at least one dimensional? And if $\pi$ is the normalised induction ... then $(\pi, 1)_{B(F)}=2$, so I would guess that over local fields too $dim\ Hom=2$.

Sounds like one of those "it's either trivial, or depends on the irrationality measure of pi being 2" questions :D

Should the error term be $1/log$?

Checkout the GAP3 package CHEVIE.

Still, heuristically the number of such $p$ should be finite: each $p$ divides $d(n_pP)$ for some $n_p$, and then there's periodicity. $\left(\frac{d(nP)}{p}\right)=1$ for all $n<n_p$ occurs with chance $2^{n_p}$. So we can estimate an upper bound for the number of "bad" primes: $\sum_n \omega(d(nP))/2^n$, where $\omega(k)$ is the number of distinct primes dividing $k$. Since $d(nP)=O(e^{cn^2})$ for some $c$, and $\omega(k)=O(\log{k})$, this sum is finite.

@Stanley Yao Xiao up to some work reconciling the definition difference, I think so: combine the main result in Reductions of Points on Elliptic Curves by Akbary et al, with the main result in Character Sums with Division Polynomials by Shparlinski and Stange.