Auction with "buy-it-now" option
Here is how I would approach the first question:


We are conditioning on the fact that you have won. Given the symmetry, this implies that your valuation $x$ is greater than your opponent's valuation $y$.
Given that you have won, you either pay the 'auction price' or $B$. If you don't pay $B$, you pay somewhere between $0$ and $p(x)$. The expected auction price is the expected value of the second highest draw, so actually $p(x)/3$ - not $p(x)/2$ as you seem to be saying (intuitively, the expected valuation of the lowest bidder must be below the unconditional expected valuation).
The probability that you pay the auction price is the probability that your opponent's valuation $y$ is below your cutoff $p(x)$. This is because your opponent will drop out of the bidding as soon as the price exceeds her valuation. $Pr(y &lt; p(x)) = p(x)$ since $y$ is uniform on $[0, 1]$.
Therefore, assuming you win, the probability that you pay $B$ is $1 - p(x)$.
Putting these claims together, I conclude that your expected payment conditional on winning equals: $(p(x)/3)p(x) + B(1 - p(x))$.
To check the plausibility of this, set $p(x) = 1$ so effectively the buy it now option disappears, and yields an expected payment of $1/3$. This is the standard result for $2$ buyers and a uniform distribution.