Using the AM-GM inequality we get

$$\begin{align*} \alpha + \beta\gamma + \gamma\delta + \delta\beta + \frac{1}{\alpha\beta^2\gamma^2\delta^2} &= \alpha + \left( \beta\gamma + \gamma\delta + \delta\beta + \frac{1}{\alpha\beta^2\gamma^2\delta^2} \right) \\ &\ge \alpha + 4\sqrt[4]{\beta\gamma \cdot \gamma\delta \cdot \delta\beta \cdot \frac{1}{\alpha\beta^2\gamma^2\delta^2}} = \alpha + \frac{4}{\sqrt[4]{\alpha}} \end{align*}$$
By putting $x = \sqrt[4]{\alpha}$ and taking in mind the given equality we have the inequality
$$x^4 + \frac{4}{x} \le 18,\ x > 0 \Leftrightarrow x^5 - 18x + 4 \le 0,\ x > 0. \quad (1)$$
Since 2 is a root of polynomial $P(x) = x^5 - 18x + 4$, we have the factorization
$$P(x) = x^{5}-18x+4 = (x-2)(x^{4}+2x^{3}+4x^{2}+8x-2),$$
and hence we have the inequality:
$$(x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2) \le 0. \quad (2)$$
If $x > 2$, then $(x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2) > 0$, and so inequality (2) is not true. Therefore, $0 < x \le 2$.
We observe that for $x=2$, we have $\alpha=16$ and (2) holds as equality. From AM - GM inequality it happens when
$$\begin{align*} \beta\gamma = \gamma\delta = \delta\beta &= \frac{1}{\alpha\beta^2\gamma^2\delta^2} \Leftrightarrow \beta = \gamma = \delta \quad \text{and} \quad \beta^2 = \frac{1}{16\beta^6} \\ \Leftrightarrow \beta = \gamma = \delta \quad \text{and} \quad \beta^8 &= \frac{1}{16} \Leftrightarrow \beta = \gamma = \delta = \frac{1}{\sqrt{2}}. \end{align*}$$
Hence the maximal possible value of $\alpha$ is 16.

Second solution.

From the AM – GM inequality we get $a + bc + cd + db + \frac{1}{ab^2c^2d^2} = \frac{a}{32} + \dots + \frac{a}{32} + bc + cd + db + \frac{1}{ab^2c^2d^2} \ge$
$$36^{36} \sqrt{\frac{a^{32}}{32^{32}} \cdot bc \cdot cd \cdot db \cdot \frac{1}{ab^2c^2d^2}} = 36^{36} \sqrt{\frac{a^{31}}{32^{32}}}$$
Hence, $\alpha^{31} \le \frac{32^{32}}{2^{36}} = 2^{124}$, and $a \le 2^4$.