I have a valid reduction below, but it is all very formal, and it's not the intuition I have for this kind of problem. When I consider, is this problem decidable/NP-Complete/etc, I frame it as a challenge: somebody says I bet BP is decidable!, and I reply Oh yeah?! Well, if that were true, then I could solve HP!. And then I show how I would do that: I suppose that BP is decidable, (i.e. I accept the challenge) and then I use that to solve HP. Granted, acknowledging this takes away some of your street cred, but it does work for me.
Here's a cheat sheet:
Show a reduction $f$ so that $HP \leq_f BP$: Hey I bet HP is decidable! Reply: Oh yeah?! Then I can solve BP like $f$
Show that Indset is NP-Complete by reduction from 3Sat, i.e. $3Sat \leq_f IndSet$: Hey I bet IndSet is easy! Reply: Oh yeah?! Well if it were easy, then I could solve 3SAT like so! I would take my arbitrary 3SAT instance $\phi$ and transform with $f$ to an IndSet instance $f(\phi)$ and then we'll let you solve $f(\phi)$ with your supposedly fast IndSet algorithm!
I hope that if these comments do not make you laugh, at least they help you internalize the notion of what a reduction is supposed to do, and how you are supposed to make one.
Now for a valid reduction. You have reduced HP to BP via a reduction $f$ (like the one I gave for your prevous question), so if somebody asked, does $M$ halt on $w$?, then you can reduce the HP instance $\langle M, w\rangle$ to the BP instance $A=f(M,w)$ (e.g. $A$ is a TM that simulates $M$ on $w$ given the blank tape), and ask your BP machine whether $A$ halts on the blank tape. So if BP were decidable, we would have an algorithm for HP, namely the one that I just described. But we don't. So BP is not decidable.