I assume you work in the Black Scholes framework. Then,
\begin{align*}
P(S_0,K,T) = Ke^{-rT}\Phi(-d_2)-S_0\Phi(-d_1),
\end{align*}
where
\begin{align*}
d_1 &= \frac{\ln\left(\frac{S_0}{K}\right)+\left(r+\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}, \\
d_2 &= \frac{\ln\left(\frac{S_0}{K}\right)+\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}= d_1-\sigma\sqrt{T}.
\end{align*}
This is an excellent webpage where you find all the derivations for the Greeks in great detail. This answers the first part of your question.
Regarding part two, we can use that $\kappa:=\frac{\partial P}{\partial K}=e^{-rT}\Phi(-d_2)$. Thus, $d_2=-\Phi^{-1}(\kappa e^{rT})$. We thus obtain a quadratic equation which we can solve for the volatility parameter $\sigma$.
\begin{align*}
\frac{\ln\left(\frac{S_0}{K}\right)+\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}} &= -\Phi^{-1}(\kappa e^{rT}) \\
\Leftrightarrow -\frac{1}{2}\sigma^2T+\Phi^{-1}(\kappa e^{rT})\sigma\sqrt{T} + \ln\left(\frac{S_0}{K}\right)+rT &= 0 \\
\Leftrightarrow \sigma^2-\frac{2}{\sqrt{T}}\Phi^{-1}(\kappa e^{rT})\sigma - \frac{2}{T}\ln\left(\frac{S_0e^{rT}}{K}\right) &= 0
\end{align*}
We then obtain as solutions
\begin{align*}
\sigma_{1,2} = \frac{1}{\sqrt{T}}\Phi^{-1}(\kappa e^{rT}) \pm\sqrt{\frac{1}{T}\Phi^{-1}(\kappa e^{rT})^2+\frac{2}{T}\ln\left(\frac{S_0e^{rT}}{K}\right)}.
\end{align*}
We may disregard one of the solutions if it's negative. For the right side, we're given the values for $r$, $S_0$ and $T$. We only need a value for $\kappa=\frac{\partial P}{\partial K}\approx \frac{P(K_1)-P(K_2)}{K_1-K_2}$. This is where we can use the two given option prices. Unlike delta, the value for $\kappa$ should be positive for a put.
However, I note that your numbers occur to be odd. With a strike of 50 and the stock price at 150, your put option is super out of the money. So, to match these prices, the volatility needs to be extremely (and unreasonably) high.
