Optimal Price and Quantity of Externality I am trying to solve the following question:
Let $h \geq 0$ represent a negative externality of a firm's production on one (representative) consumer. The consumer has a quasi-linear utility function and attaches a utility of $\phi(h) = -2h^2$ to the externality. The firm's profit function is $\pi(h)=120-2(h-10)^2 $. Suppose the consumer has the property right concerning $h$ and can sell the right to produce a quantity $h$ at some price $P$. The consumer's utility function from some combination $(h,P)$ is $u(h,P)= \phi(h) + P$. The firm's profit function is $\Pi(h,P)=\pi(h)-P$.

Suppose the firm can make the consumer a take-it-or-leave-it offer for an externality level $h$ at price $P$. If the consumer rejects the firm's offer the firm does not produce so the consumer's utility is 0. Compute the firm's optimal offer $(h_p^f,P^f)$ to the consumer.
Suppose the consumer can make the firm a take-it-or-leave-it offer for an externality level $h$ at price $P$. If the firm rejects the offer it cannot produce so the firm's profit is 0. Compute the consumer's optimal offer $(h_p^c,P^c)$ to the firm.

Edit: Approach using the hint from Herr K.

The consumer can reject the offer giving him a guaranteed utility of 0. Maximize the firm's profit subject to this constraint, $u(h,P)= \phi(h) + P = 0$. Solve this using Lagrangian method.
$$ Lagrangian = objective function + constraint  $$
$$\mathcal{L}(h,P,\lambda) = \Pi(h,P) + \lambda(\phi(h)+P)$$
$$\mathcal{L}(h,P,\lambda) = \pi(h) - P + \lambda(-2h^2+P)$$
$$\mathcal{L}(h,P,\lambda) = 120-2(h-10)^2 - P + \lambda(-2h^2+P)$$
$$\frac{\partial\mathcal{L}}{\partial h} = -4(h-10)-4h\lambda = 0 \rightarrow \lambda = \frac{-4h+40}{4h}$$
$$\frac{\partial\mathcal{L}}{\partial P} = -1 + \lambda = 0 \rightarrow \lambda = 1$$
$$\frac{\partial\mathcal{L}}{\partial \lambda} = -2h^2+P = 0$$

When we set both $\lambda$ equal to each other we get $1=\frac{-4h+40}{4h} \rightarrow h=5$. Plugging this into $\frac{\partial\mathcal{L}}{\partial \lambda}$ we get $-2(5)^2+P=0 \rightarrow 50=P$
Thus $h_p^f = 5$ and $P^f= 50$

We do the same thing but this time the objective function is the consumer's utility function $u(h,P)= \phi(h) + P$ and the constraint is the firm's profit function $\Pi(h,P)=\pi(h)-P$ = 0.

$$ Lagrangian = objective function + constraint  $$
$$\mathcal{L}(h,P,\lambda) = \phi(h)+P + \lambda(\Pi(h,P))$$
$$\mathcal{L}(h,P,\lambda) = -2h^2+P + \lambda(\pi(h) - P)$$
$$\mathcal{L}(h,P,\lambda) = -2h^2+P + \lambda(120-2(h-10)^2 - P)$$
$$\frac{\partial\mathcal{L}}{\partial h} = -4h -4\lambda(h-10) = 0 \rightarrow \lambda = -\frac{h}{h-10}$$
$$\frac{\partial\mathcal{L}}{\partial P} = 1 -\lambda = 0 \rightarrow \lambda = 1$$
$$\frac{\partial\mathcal{L}}{\partial \lambda} = 120-2(h-10)^2 - P = 0$$
When we set both $\lambda$ equal to each other we get $1=\frac{h}{h-10} \rightarrow 2h=5$. Plugging this into $\frac{\partial\mathcal{L}}{\partial \lambda}$ we get $120-2((5)-10)^2 - P=0 \rightarrow 70=P$
Thus $h_p^c = 5$ and $P^c= 70$
Your steps look correct. There is one small typo in the first $\frac{\partial \mathcal L}{\partial h}$: the term $-4h(h-10)$ should have been $-4(h-10)$.
The results also look reasonable: Regardless of who makes the offer, the Pareto optimal level of $h$ is produced. Given how the bargaining procedures are structured, whoever gets to make the offer gets to keep the surplus of $20$.

P.S. There is an issue in part 2 of the question. It says: &quot;If the firm rejects the offer it cannot produce so the firm's profit is 0.&quot; This statement is a contradiction. If the firm cannot produce ($h=0$), then profit would be $-80&lt;0$; if its profit is $0$, then it must produce some positive amount of $h$. But this lack of rigor is on whoever wrote the question, not you.