OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $\pu{90.08 gmol-1}$) and potassium hydroxide (MW: $\pu{50.11 gmol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%(w/w) \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$. Thus, weight ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Simillarly, since OP has only $88\%(w/w)$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid. Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}= \pu{311.7 g}$$

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88} = \pu{511.8 g}$$

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%(w/w)$. Otherwise, additional information such as density is needed.

OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$. Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid. Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

\begin{align} m_{\ce{KOH}} &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90} = \pu{311.7 g}\\ m_{\mathrm{LA}} &= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88} = \pu{511.8 g} \end{align}

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.