How to calculate the concentration of conjugate acid from given pH and pKa values?
To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.

Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:

$$\ce{B + H2O&lt;=&gt;BH +OH-}$$

So the equilibrium expression of the reaction is given by the equation:

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:

$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$

$$-\log K_\mathrm{a}=8.1$$

$$\log K_\mathrm{a}=-8.1$$

$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$

Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:

$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$

$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$

For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:

$$14=8.1+\mathrm{p}\ce{OH}$$

$$\mathrm{pOH}=5.9$$
and
$$-\log{[\ce{OH-}]}=5.9$$

$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$

Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then: 

$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$

Using the knowledge we have gained from the problem, we can now solve (finally!):

$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$

$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$

$$x=0.1-x$$

$$2x=0.1$$

$$x=0.05$$

So:

$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$

Try using this process to solve for the $\mathrm{pH}$ of 7.5.